OP Malhotra Class 10 Maths Solutions Chapter 19 Histogram and Ogive Exercise 19 (D)

S Chand Class 10 ICSE Maths Solutions Chapter 19 Histogram and Ogive Ex 19(d)

 

Question 1. Find out the median by plotting the following in the form of an ogive, where 'm' denotes the mid-mark of a class :

Answer: First, we convert the mid-marks into class intervals and then find the cumulative frequency (c.f.). The mid-mark 'm' helps us define the class boundaries; for example, 12.5 is the mid-mark for the class 10-15.We then plot the points (upper class boundary, cumulative frequency) on a graph: (15, 2), (20, 24), (25, 43), (30, 57), (35, 60), (40, 64), (45, 70), (50, 71), (55, 72). We join these points with a free hand to create the ogive curve.Here, the total number of items, \( n \), is 72, which is an even number. To find the median, we calculate the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{72}{2} = 36 \) Now, locate 36 on the y-axis (cumulative frequency axis). From this point, draw a horizontal line towards the ogive curve. Let this line meet the curve at point P. From point P, draw a perpendicular line downwards to the x-axis. The point where this perpendicular line meets the x-axis, let's call it M, represents the median. From the graph, the median is approximately 23.5. Drawing an ogive is a visual way to understand the distribution of data.In simple words: First, list the data with how many times each value appears (frequency) and a running total (cumulative frequency). Then, draw a graph with cumulative frequency on the vertical line and class marks on the horizontal line. Find the middle spot on the vertical line, draw a straight line to the graph, and then drop a line down to find the median.

🎯 Exam Tip: When drawing an ogive, ensure your cumulative frequency values are plotted against the upper class boundaries. For median calculation, always use the formula \( \frac{N}{2} \) or \( \frac{N+1}{2} \) to find the position before locating on the graph.

 

Question 2. From the data given below draw an ogive and find the values of median and quartiles from the graph.

Answer: We plot the given points directly on the graph since "Years under" directly gives the upper class boundaries and "Number of persons" gives the cumulative frequency. The points are: (10, 12), (20, 31), (30, 44), (40, 55), (50, 65), and (60, 70). Connect these points with a free-hand curve to form the ogive. This curve visually represents the distribution of ages.(i) Here, the total number of persons, \( n \), is 70. The median term is the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{70}{2} = 35 \)th term. To find the median, locate 35 on the y-axis (A). Draw a horizontal line from A to meet the ogive at P. From P, draw a perpendicular line PL to the x-axis. The value at L on the x-axis is the median. From the graph, L is approximately 23 years. Thus, the median is 23 years. (ii) For the First Quartile (Q1): Q1 is the \( \frac{n}{4} \)th term. \( \frac{n}{4} = \frac{70}{4} = 17.5 \)th term. Locate 17.5 on the y-axis (B). Draw a line parallel to the x-axis from B to meet the curve at Q. From Q, draw a perpendicular line QM to the x-axis. From the graph, M is approximately 13 years. Thus, the first quartile (Q1) is 13 years. (iii) For the Third Quartile (Q3): Q3 is the \( \frac{3n}{4} \)th term. \( \frac{3 \times 70}{4} = \frac{210}{4} = 52.5 \)th term. Locate 52.5 on the y-axis (C). Draw a line parallel to the x-axis from C to meet the curve at R. From R, draw a perpendicular line RN to the x-axis. From the graph, N is approximately 37 years. Thus, the third quartile (Q3) is 37 years.In simple words: After drawing the curve, to find the median, look for the middle value on the y-axis (total number of people divided by two). Go across to the curve, then down to the x-axis. This gives you the median age. For quartiles, do the same but for a quarter of the total (for Q1) and three-quarters of the total (for Q3).

🎯 Exam Tip: Always clearly label the axes and points (P, Q, R, L, M, N) on your ogive graph when estimating median and quartiles to ensure full marks.

 

Question 3. From the following find out the median with the help of ogive curve :

Answer: The data "Profit per shop less than" means these are the upper limits for the profits, and "No. of shops" are the cumulative frequencies. So, we plot the points: (10, 12), (20, 30), (30, 57), (40, 77), (50, 94) and (60, 100) on the graph. Then, we join these points with a free-hand curve to create the ogive. This curve helps visualize the distribution of profit per shop.Here, the total number of shops, \( n \), is 100. The median term is the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{100}{2} = 50 \)th term. To find the median, locate 50 on the y-axis (A). Draw a horizontal line from A to meet the ogive at P. From P, draw a perpendicular line PL to the x-axis. The value at L on the x-axis is the median. From the graph, L is approximately 27.8. So, the median profit is about 28 (approx). The median value shows the point where half the shops earn less and half earn more profit.In simple words: Draw a graph using the "Profit less than" numbers and the "No. of shops" numbers. Find the middle point of all shops (which is 50). Go across from this point on the left side to your graph line, then go straight down to find the median profit.

🎯 Exam Tip: When dealing with "less than" data, the given frequencies are already cumulative frequencies, and the "less than" values are the upper class boundaries for plotting the ogive.

 

Question 4. The following are the marks obtained by 50 students in Statistics :

Answer: We will plot the points: (10, 4), (20, 10), (30, 30), (40, 40), (50, 47) and (60, 50) on the graph. Then, we join these points with a free-hand curve to construct the ogive. This curve shows how the marks are distributed among the students.Here, the total number of students, \( n \), is 50. The median term is the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{50}{2} = 25 \)th term. To find the median, locate 25 on the y-axis. Draw a horizontal line from this point to meet the ogive at P. From P, draw a perpendicular line to the x-axis, meeting it at L. The value at L on the x-axis is the median. From the graph, L is approximately 27.8. So, the median marks are about 28. This means half the students scored less than 28 marks, and half scored more.In simple words: First, draw the ogive graph with marks on the bottom and the count of students on the left. Find the middle student (student number 25). Draw a line across from student 25 to the curved line, then draw a line straight down to the marks line. The mark you land on is the median.

🎯 Exam Tip: Remember to always use the cumulative frequency for the y-axis when constructing an ogive curve. "Marks less than" directly provides the upper class boundaries and the cumulative frequencies.

 

Question 5. 100 pupils in a school have heights as tabulated below :

Answer: First, we convert the given inclusive class intervals into exclusive ones to ensure continuity for the ogive, and then calculate the cumulative frequency (c.f.). The exclusive intervals help represent the data more accurately on a continuous scale.Next, we plot the points using the upper class boundaries and the cumulative frequencies: (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92) and (180.5, 100). We join these points with a free-hand curve to form the ogive. This graph helps visualize the distribution of heights.Here, the total number of pupils, \( n \), is 100. The median term is the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{100}{2} = 50 \)th term. To find the median, locate 50 on the y-axis. Draw a horizontal line from this point to meet the ogive at P. From P, draw a perpendicular line PL to the x-axis. The value at L on the x-axis is the median height. From the graph, L is approximately 147.8 cm, which can be rounded to 148 cm. The median height indicates that half of the pupils are shorter than 148 cm and half are taller.In simple words: First, rewrite the height groups so they flow smoothly, then add up the number of students as you go (cumulative frequency). Plot these points on a graph. Find the middle student (student number 50). Go across from student 50 on the left to the curve, then down to the height line. This height is the median.

🎯 Exam Tip: When given inclusive class intervals, always convert them to exclusive intervals before plotting the ogive to ensure the curve is continuous and accurate for estimation.

 

Question 6. Draw an ogive curve from the following data and find out (a) the median wage and (b) number of workers earning less than Rs. 55 per week :

Answer: First, we create a cumulative frequency table from the given data. This helps us see the running total of workers for each wage group.Next, we plot the points using the upper class boundaries and the cumulative frequencies: (20, 40), (40, 91), (60, 151), (80, 189) and (100, 196). We join these points with a free-hand curve to form the ogive. This curve shows the distribution of weekly wages.Here, the total number of workers, \( n \), is 196. (i) To find the median wage: The median term is the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{196}{2} = 98 \)th term. Locate 98 on the y-axis. Draw a horizontal line from this point to meet the ogive at P. From P, draw a perpendicular line PL to the x-axis. The value at L on the x-axis is the median. From the graph, L is approximately 42.3. So, the median wage is Rs. 42.3. This means half the workers earn less than Rs. 42.3, and half earn more. (ii) To find the number of workers earning less than Rs. 55 per week: Locate 55 on the x-axis. Draw a perpendicular line from this point to meet the ogive at Q. From Q, draw a horizontal line to meet the y-axis at B. The value at B on the y-axis indicates the number of workers earning less than Rs. 55. From the graph, B is approximately 136. So, the number of workers earning less than Rs. 55 per week is 136.In simple words: First, draw a graph using wages on the bottom and the total count of workers on the left. To find the median wage, locate the middle worker (number 98) on the left side, go across to the graph line, then straight down to the wage line. For workers earning less than Rs. 55, find Rs. 55 on the wage line, go straight up to the graph line, then across to the left to find the number of workers.

🎯 Exam Tip: When finding values *from* the graph, such as median or the number of workers earning less than a certain amount, always explicitly state the lines drawn from the axes to the curve and back for clarity.

 

Question 7. From the data given below, draw a cumulative frequency curve and locate the median and quartiles graphically :

Answer: First, we rewrite the data in ascending order of income and then create a cumulative frequency table. This ordered table is essential for drawing an accurate ogive.Now, we plot the points using the upper class boundaries and the cumulative frequencies: (10, 50), (20, 150), (30, 300), (40, 500), (50, 700) and (60, 800) on the graph. We join these points with a free-hand curve to create the ogive. This curve shows the distribution of income among the persons.Here, the total number of persons, \( n \), is 800. The median term is the \( \frac{n}{2} \)th term. \( \frac{n}{2} = \frac{800}{2} = 400 \)th term. To find the median, locate 400 on the y-axis. Draw a horizontal line from this point to meet the ogive at P. From P, draw a perpendicular line PL to the x-axis. The value at L on the x-axis is the median. From the graph, L is approximately 35. So, the median income is Rs. 35. This means half the people earn less than Rs. 35, and half earn more. (i) For the First Quartile (Q1): Q1 is the \( \frac{n}{4} \)th term. \( \frac{n}{4} = \frac{800}{4} = 200 \)th term. Locate 200 on the y-axis. Draw a line parallel to the x-axis from this point to meet the curve at Q. From Q, draw a perpendicular line QM to the x-axis. From the graph, M is approximately 23. So, the first quartile (Q1) is Rs. 23. (ii) For the Third Quartile (Q3): Q3 is the \( \frac{3n}{4} \)th term. \( \frac{3 \times 800}{4} = \frac{2400}{4} = 600 \)th term. Locate 600 on the y-axis. Draw a line parallel to the x-axis from this point to meet the curve at R. From R, draw a perpendicular line RN to the x-axis. From the graph, N is approximately 45. So, the third quartile (Q3) is Rs. 45.In simple words: First, arrange the income data from smallest to largest and create a running total (cumulative frequency). Plot these points on a graph. To find the median, locate the middle person (person number 400) on the left side, go across to the graph line, then down to the income line. For Q1, use person number 200, and for Q3, use person number 600, following the same steps.

🎯 Exam Tip: When the data is not initially ordered, always rearrange it in ascending order before constructing the cumulative frequency table and plotting the ogive to ensure correct median and quartile calculations.

 

Question 8. Draw the ogive for the following frequency distribution. Estimate the median from your graph.

Answer: First, we convert the inclusive class intervals into exclusive ones and then create a cumulative frequency table. This step is crucial for accurate plotting of the ogive.Now, we plot the points using the upper class boundaries and the cumulative frequencies: (10.5, 2), (20.5, 7), (30.5, 14), (40.5, 26), (50.5, 34), (60.5, 41) and (70.5, 45). We join these points with a free-hand curve to form the ogive. This curve visually represents the frequency distribution.Here, the total number of items, \( n \), is 45, which is an odd number. The median term is the \( \frac{n+1}{2} \)th term. \( \frac{n+1}{2} = \frac{45+1}{2} = \frac{46}{2} = 23 \)rd term. To find the median, locate 23 on the y-axis. Draw a horizontal line from this point to meet the ogive at P. From P, draw a perpendicular line PL to the x-axis. The value at L on the x-axis is the median. From the graph, L is approximately 38.1. So, the median is approximately 38.1. The median represents the central value of the distribution.In simple words: First, rewrite the class groups to be continuous (exclusive form) and calculate the running total of frequencies. Then, plot these cumulative frequency points on a graph. Find the middle point of all data entries (the 23rd one). From this point on the left side, draw a line across to your graph curve, then draw a line straight down to the bottom line. The number you hit is the median.

🎯 Exam Tip: For odd 'n' values, the median term is simply \( \frac{n+1}{2} \)th term. For even 'n' values, it's the average of the \( \frac{n}{2} \)th and \( (\frac{n}{2}+1) \)th terms, but for an ogive, you typically still find the single point corresponding to the \( \frac{n}{2} \)th cumulative frequency.

 

Question 9. Attempt this question on graph paper.
Age (yrs) 5-15 15-25 25-35 35-45 45-55 55-65 65-75
No. of casualties 6 10 15 13 24 8 7
due to accidents
(i) Construct the 'less than' cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.
(ii) From your graph determine.
(b) the upper quartile
Answer: First, let's make a cumulative frequency table for the given data:

Now, we plot the points for the ogive: (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65, 76) and (75, 83) on a graph paper and join them with a free hand to create a 'less than' ogive. The graph will show how the cumulative frequency changes with age. Here, the total number of casualties \(n = 83\).

(i) For the median, since \(n = 83\) (odd), the median term is \( \frac{n+1}{2} = \frac{83+1}{2} = 42^{nd} \) term.
From 41.5 on the y-axis, draw a horizontal line that meets the curve at P. From P, draw a perpendicular line down to the x-axis, which meets it at M.
M is the median, and its value is 43 years.

(ii) For the upper quartile (Q3), the term is \( \frac{3n}{4} = \frac{3 \times 83}{4} = \frac{249}{4} = 62.25^{th} \) term.
From 62.25 on the y-axis, draw a horizontal line that meets the curve at Q. From Q, draw a perpendicular line down to the x-axis, which meets it at N.
N represents Q3, and its value is 52 years.
In simple words: First, we list all the data and count how many items there are. Then we draw a special graph called an ogive, which shows how the total number of casualties adds up. We use this graph to find the middle value (median) and the top quarter value (upper quartile) by locating their positions on the y-axis and then reading the corresponding value on the x-axis.

🎯 Exam Tip: Remember to always check if 'n' (total frequency) is odd or even when calculating the median term. For ogives, be precise with your graph plotting and reading points from the axes to ensure accurate median and quartile estimations.

 

Question 10. 100 students in an examination are as follows :
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
No. of students 10 20 22 40 55 75 80 58 28 12
Draw an ogive and from it determine:
(i) the median mark,
(ii) pass marks if 80% of the students pass examination.
Answer: First, let's create a cumulative frequency table from the given data:

Now, plot the points (10, 10), (20, 30), (30, 52), (40, 92), (50, 147), (60, 222), (70, 302), (80, 360), (90, 388), (100, 400) on a graph. Join these points with a free hand to create the ogive as shown below:

(i) Here, the total number of students \(n = 400\). The median term is \( \frac{n}{2} = \frac{400}{2} = 200^{th} \) term.
From 200 (point A) on the y-axis, draw a horizontal line to meet the ogive at P. From P, draw a perpendicular line to the x-axis, meeting it at M.
M is the median, and its value is 57.5 marks.
(ii) For students to pass, 80% of them must pass. This means \( 80\% \) of \( 400 = \frac{80}{100} \times 400 = 320 \) students.
The number of students who did not pass is \( 400 - 320 = 80 \).
From the point 80 on the y-axis, draw a line parallel to the x-axis, meeting the curve at Q. From Q, draw a perpendicular line to the x-axis, meeting it at N. This point N represents the pass marks.
The pass marks are 36.
In simple words: We first make a list that shows how many students got marks up to a certain point (cumulative frequency). Then, we draw a graph from this list. To find the median, we find the middle student on the y-axis and see what mark they got on the x-axis. To find the pass mark, we work backwards: if 80% pass, then 20% fail. We find the mark below which 20% of students fall, and that mark is the passing grade.

🎯 Exam Tip: When using an ogive for pass/fail cutoffs, remember that "80% pass" means 20% *fail*. You need to find the mark for the cumulative frequency corresponding to the failed percentage, which will be the minimum passing mark.

 

Question 11. Draw an ogive for the following frequency distribution. Use your ogive to estimate
(i) the median
(ii) the number of students who obtained more than 75% marks (use square paper to solve this question).
Marks 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
No. of students 5 9 16 22 26 18 11 6 4 3
Answer: First, let's convert the given class intervals to exclusive form and create a cumulative frequency table:

Now plot the points (9.5, 5), (19.5, 14), (29.5, 30), (39.5, 52), (49.5, 78), (59.5, 96), (69.5, 107), (79.5, 113), (89.5, 117), and (99.5, 120) on the graph. Join them with a free hand to create an ogive as shown below:

Here, the total number of students \(n = 120\).
(i) Median: The median term is \( \frac{n}{2} = \frac{120}{2} = 60^{th} \) term.
From 60 on the y-axis, draw a horizontal line to meet the curve at P. From P, draw a perpendicular line to the x-axis, meeting it at M.
M is the median, and its value is 43 marks.

(ii) Number of students who obtained more than 75% marks.
Marks for 75% are 75.
From 75 on the x-axis, draw a vertical line to meet the curve at Q. From Q, draw a horizontal line to meet the y-axis at N.
Now, \(Q\) on the y-axis is 110. This means 110 students scored up to 75 marks. Students scoring more than 75% means those above 75 marks.
Total no. of students getting more than 75% marks = \(120 - 110 = 10\) students.
In simple words: First, we change the mark groups to be clear, then count up the students to get a running total. We draw a line graph (ogive) from this. To find the median, we find the middle student on the y-axis and see their mark. To find how many got more than 75%, we find 75 marks on the x-axis, go up to the graph, then across to the y-axis to see how many students scored *up to* 75. Then we subtract that number from the total students to find how many scored *more* than 75.

🎯 Exam Tip: When dealing with "more than" questions on an ogive, remember you're looking for the difference between the total number of students and the number of students *less than or equal to* that specific mark.

 

Question 12. In a public collection towards the erection of a memorial, 1000 people contributed sums of money varying from Rs. 1 to Rs. 100 (in units of Re. 1). The following table gives the frequency distribution of contribution :
Contribution in Rs.) 1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100
No. of people 30 60 80 170 200 180 140 70 40 30
Using a suitable scale, draw on graph paper an ogive (cumulative frequency graph) and use it to answer the following:
(i) Estimate the median.
(ii) If it is agreed to allow only those who contributed Rs. 45 or more to attend the unveiling ceremony, what percentage would attend ?
Answer: First, we rewrite the data with exclusive class intervals and create a cumulative frequency table:

Now, plot the points (10.5, 30), (20.5, 90), (30.5, 170), (40.5, 340), (50.5, 540), (60.5, 720), (70.5, 860), (80.5, 930), (90.5, 970) and (100.5, 1000) on the graph. Join them with a free hand to create an ogive as shown below:

Here, the total number of people \(n = 1000\).
(i) Median: The median term is \( \frac{n}{2} = \frac{1000}{2} = 500^{th} \) term.
From 500 on the y-axis, draw a horizontal line to meet the curve at P. From P, draw a perpendicular line to the x-axis, meeting it at L.
L is the median, which is approximately Rs. 49.
(ii) For people who contributed Rs. 45 or more to attend: From Rs. 45 on the x-axis, draw a perpendicular line to meet the curve at Q. From Q, draw a horizontal line to meet the y-axis at B. B is 440. This means 440 people contributed less than Rs. 45.
The number of people who contributed Rs. 45 or more is \( 1000 - 440 = 560 \).
The percentage of people who would attend is \( \frac{560 \times 100}{1000} = 56\% \).
In simple words: We organize the money collected into groups and count how many people gave up to a certain amount. We draw a graph from this. To find the median, we locate the middle person on the 'number of people' line and see how much money they gave. To find the percentage who gave Rs. 45 or more, we find Rs. 45 on the 'money' line, see how many people gave *less* than that, and then subtract that from the total to find how many gave *more*. Finally, we turn that number into a percentage.

🎯 Exam Tip: When using an ogive, remember that "Rs. 45 or more" means finding the cumulative frequency at Rs. 45 and subtracting it from the total frequency. This will give you the count of items in the upper tail of the distribution.

Self-Evaluation and Revision (LATEST ICSE QUESTIONS)

 

Question 1. Draw a histogram to represent the following data :
Pocket money (in Rs.) 150-200 200-250 250-300 300-350 350-400
No. of students 10 5 7 4 3
Answer: The frequency distribution table is as follows:

Represent the pocket money along the x-axis and the number of students along the y-axis. Then, draw the histogram as shown:

The frequency polygon is obtained by joining the consecutive mid-points of the upper sides (tops) of the adjacent rectangles of the histogram by means of line segments. Now, join the two vertices A and B of the highest rectangle, diagonally to the upper corners C and D of the adjacent rectangles on either side of the highest rectangles, by line segments AC and BD. Let AC and BD intersect at L. From L, draw the perpendicular LM on the horizontal line (the x-axis). Then, the abscissa of the point M, i.e., OM = 172.5, determines the mode. Thus, the required mode is Rs. 172.50.
In simple words: We draw a bar graph (histogram) where the width of each bar is the range of pocket money and the height is the number of students. The tallest bar shows the most common pocket money range. To find the mode, we draw lines from the top corners of the tallest bar to the adjacent bars. Where these lines cross, we drop a line to the bottom axis, and that point tells us the most frequent (modal) pocket money amount.

🎯 Exam Tip: When constructing a histogram, ensure that there are no gaps between adjacent bars, as this represents continuous data. The mode is always found from the tallest rectangle in the histogram by drawing the diagonals.

 

Question 2. Using a graph paper, draw an Ogive for the following distribution which shows a record of the weight in kilogram of 200 students.
Weight 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80
Frequency 5 17 22 45 51 31 20 9
Use the ogive to estimate the following :
(i) The percentage of students weighing 55 kg or more;
(ii) The weight above which the heaviest 30% of the students fall;
(iii) The number of students who are
(1) under weight
(2) over weight, if 55.70 kg is considered as the standard weight.
Answer: First, let's create a cumulative frequency table:

Now, plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph. Join them with a free hand to create an ogive as shown below:

From the graph:
(i) To find the percentage of students weighing 55 kg or more: Draw a vertical line from 55 kg on the x-axis to meet the ogive at A. From A, draw a horizontal line to the y-axis, meeting it at D. The value at D is 44. This means 44 students weigh less than 55 kg.
The number of students weighing 55 kg or more is \( 200 - 44 = 156 \).
Percentage = \( \frac{156 \times 100}{200} = 78\% \).
(ii) To find the weight above which the heaviest 30% of students fall: 30% of the total students (200) is \( \frac{30}{100} \times 200 = 60 \) students. These 60 students are the heaviest. This means we are looking for the weight at which \( 200 - 60 = 140 \) students fall below this weight. From 140 on the y-axis, draw a horizontal line to the ogive at P. From P, draw a vertical line to the x-axis. The weight is 65 kg. So, the heaviest 30% of students weigh above 65 kg.
(iii) Number of students who are under weight (below 55.70 kg) and over weight (above 55.70 kg) if 55.70 kg is the standard weight:
From 55.70 kg on the x-axis, draw a vertical line up to the ogive and then a horizontal line to the y-axis. This point is 47 (approximately). So, the number of students under weight (below 55.70 kg) is 47.
The number of students over weight (above 55.70 kg) is \( 200 - 47 = 153 \).
In simple words: First, we list the weights and count how many students fall into each weight group (cumulative frequency), then draw a graph. For students weighing 55 kg or more, we find 55 kg on the bottom line, go up to the graph, then across to the side line to see how many weigh *less* than 55 kg. We subtract this from the total to get the number weighing *more*. For the heaviest 30%, we count down 30% from the total students on the side line and see what weight they correspond to. For specific standard weight, we find that weight on the bottom line and see how many students are above or below it.

🎯 Exam Tip: When estimating values from an ogive, use a ruler to draw precise lines from the axes to the curve and back. For percentages like "heaviest 30%", remember to find the corresponding cumulative frequency from the *top* of the distribution (total frequency minus 30% of total) and then read the weight from the x-axis.

 

Question 3. If the mean of the following distribution is 7.5 find the missing frequency f:
Variable 5 6 7 8 9 10 11 12
Frequency 20 17 f 10 8 6 7 6
Answer: First, we construct a table to calculate the sum of frequencies (\( \Sigma f \)) and the sum of \(f \times x\) (\( \Sigma fx \)):

The formula for the mean (\( \bar{x} \)) is \( \bar{x} = \frac{\Sigma fx}{\Sigma f} \).
Given that the mean is 7.5, we can set up the equation:
\( 7.5 = \frac{563 + 7f}{74 + f} \)
Now, we solve for \(f\):
\( 7.5 (74 + f) = 563 + 7f \)
\( 555 + 7.5f = 563 + 7f \)
\( 7.5f - 7f = 563 - 555 \)
\( 0.5f = 8 \)
\( f = \frac{8}{0.5} \)
\( f = \frac{8 \times 10}{5} \)
\( f = \frac{8 \times 2}{1} \)
\( f = 16 \)
The missing frequency is 16.
In simple words: We are given the mean (average) of some numbers and a list of how often each number appears, but one count is missing. We set up a table to multiply each number by its count. Then, we use the mean formula, which is total of (number × count) divided by total count. We put in all the numbers we know and solve the equation to find the missing count.

🎯 Exam Tip: Always remember to sum both the frequencies (\( \Sigma f \)) and the product of frequency and variable (\( \Sigma fx \)) accurately. A small calculation error here can lead to a completely wrong value for the missing frequency.

 

Question 4. The median of the following observation 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. Find x.
Answer: The total number of observations, \(n\), is 9. Since 9 is an odd number, the median is the \( \frac { n+ 1 }{ 2 } \)th term. So, the median is the \( \frac { 9 + 1 }{ 2 } = 5 \)th term. The 5th term in the given arranged data is \(x + 4\). We are told that the median is 24. This means \( x + 4 = 24 \). To find \(x\), we subtract 4 from 24, so \( x = 24 - 4 \). Therefore, \(x = 20\).
In simple words: We have 9 numbers in order, and the middle number (the 5th one) is \(x+4\). We know this middle number is 24, so we just figure out what \(x\) must be.

🎯 Exam Tip: Always count the number of observations (n) first to determine if it's odd or even, as this changes the median formula slightly. For ordered data, ensure no data points are missed.

 

Question 5. Find the mean of the following distribution:

Answer: To find the mean of a grouped frequency distribution, we first need to calculate the mid-value for each class interval. Then, we multiply each mid-value by its frequency to get \(f \cdot x\). Finally, we sum all \(f \cdot x\) values and divide by the total sum of frequencies.

Now, we calculate the mean:
\( \bar{x} = \frac{\sum f x}{\sum f} = \frac{2480}{50} = 49.6 \)
In simple words: We find the middle point of each group, multiply it by how many times it appears, add all those up, and then divide by the total count of numbers. This gives us the average value.

🎯 Exam Tip: Clearly label your columns in the frequency table. Calculate mid-values accurately, as any error here will affect the final mean.

 

Question 6. The daily wages of 160 workers in a building project are given below :

Using a graph, draw an ogive for the above distribution. Use your ogive to estimate :
(i) the median wage of the workers;
(ii) the upper quartile wage of the workers;
(iii) the lower quartile wage of the workers;
(iv) the percentage of workers who earn more than Rs. 45 a day.
Answer: First, we need to create a cumulative frequency table from the given data. This table will help us plot the ogive. An ogive graph helps us find values like median and quartiles by looking at the cumulative frequencies. These graphs are quite useful in data analysis to understand data distribution.

We plot the points (10, 12), (20, 32), (30, 62), (40, 100), (50, 124), (60, 140), (70, 152) and (80, 160) on the graph and join them with a free hand curve to get the ogive. Here, the total number of workers \(n = 160\).
(i) Median: \( \frac{n}{2} = \frac{160}{2} = 80 \)th term. From 80 on the y-axis, draw a horizontal line that meets the ogive at point P. From P, draw a perpendicular line to the x-axis, meeting it at L. The value of L is 35. Therefore, the median wage is Rs. 35.
(ii) Upper quartile (\(Q_3\)): \( \frac{3n}{4} = \frac{3 \times 160}{4} = 120 \)th term. From 120 on the y-axis, draw a horizontal line that meets the ogive at point Q. From Q, draw a perpendicular line to the x-axis, meeting it at M. The value of M is 47.50. Therefore, the upper quartile wage is Rs. 47.50.
(iii) Lower quartile (\(Q_1\)): \( \frac{n}{4} = \frac{160}{4} = 40 \)th term. From 40 on the y-axis, draw a horizontal line that meets the ogive at point R. From R, draw a perpendicular line to the x-axis, meeting it at N. The value of N is 22.5. Therefore, the lower quartile wage is Rs. 22.5.
(iv) Percentage of workers earning more than Rs. 45 a day: From Rs. 45 on the x-axis, draw a perpendicular line that meets the ogive at point B. From B, draw a horizontal line that meets the y-axis at C. The value of C is 112. This means 112 workers earn less than or equal to Rs. 45. The total number of workers is 160. So, the number of workers earning more than Rs. 45 is \( 160 - 112 = 48 \). The percentage is \( \frac{48 \times 100}{160} = 30\% \).
In simple words: We used the graph to find different income levels for the workers. The median income is Rs. 35, meaning half the workers earn less and half earn more. The lowest 25% earn up to Rs. 22.5, and the highest 25% earn more than Rs. 47.50. Also, 30% of the workers earn more than Rs. 45 per day.

🎯 Exam Tip: When drawing ogives, ensure your scale is consistent and clearly marked. Use a ruler and pencil for accuracy when drawing lines to find median and quartiles, as small inaccuracies can lead to incorrect answers. Show all construction lines on your graph.

 

Question 7. Find the mean of the following distribution:

Answer: To find the mean of this grouped frequency distribution, we'll calculate the mid-value for each class, multiply it by its frequency, sum these products, and then divide by the total frequency. This method is standard for finding the average of data organized in groups.

\( \text{Mean} = \frac{\sum f x}{\sum f} = \frac{985}{41} = 24.02 \)
In simple words: We calculated the average value by taking the middle point of each group, multiplying it by how many data points are in that group, adding up all these results, and then dividing by the total number of data points. The average turned out to be about 24.02.

🎯 Exam Tip: Remember to calculate mid-values carefully. A common mistake is using the lower or upper boundary instead of the midpoint, which will lead to an incorrect mean.

 

Question 8. The table below shows the distributions of the scores obtained by 120 shooters in a shooting competition. Using a graph sheet, draw in ogive for the distributions.
Use your ogive to estimate : (i) The median (ii) The inter quartile range (iii) The number of shooters who obtained more than 75% scores.

Answer: First, we'll build a cumulative frequency table, which helps us track the running total of shooters up to a certain score. This table is essential for drawing an ogive, which visually represents the cumulative distribution. The ogive allows us to estimate median, quartiles, and other specific values easily from the graph.We plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), and (100, 120) on the graph and join them with a free hand curve to get the ogive. Here, \(n = 120\).
(i) Median: The total number of students \(n = 120\). The median position is \( \frac{n}{2} = \frac{120}{2} = 60 \)th term. From 60 on the y-axis, draw a horizontal line to meet the ogive at P. From P, draw a perpendicular line to the x-axis, meeting it at L. The abscissa of L is 57. Thus, the median score is 57.
(ii) Interquartile range: Lower quartile (\(Q_1\)): \( \frac{n}{4} = \frac{120}{4} = 30 \)th term. From 30 on the y-axis, draw a horizontal line to meet the ogive at Q. From Q, draw a perpendicular line to the x-axis, meeting it at M. The abscissa of M is 30. So, \(Q_1 = 30\).
Upper quartile (\(Q_3\)): \( \frac{3n}{4} = \frac{3 \times 120}{4} = 90 \)th term. From 90 on the y-axis, draw a horizontal line to meet the ogive at R. From R, draw a perpendicular line to the x-axis, meeting it at N. The abscissa of N is 57. So, \(Q_3 = 57\).
Interquartile range \( = Q_3 - Q_1 = 57 - 30 = 27 \). This range tells us how spread out the middle 50% of the scores are.
(iii) Number of shooters scoring more than 75% marks: From 75 on the x-axis, draw a perpendicular line that meets the curve at B. From B, draw a line parallel to the x-axis, meeting the y-axis at C. The ordinate of C is 110. This means 110 shooters scored 75 marks or less. The total number of shooters is 120. Therefore, the number of shooters getting more than 75% marks is \( 120 - 110 = 10 \).
In simple words: We used the graph to find out some key things about the shooter's scores. The middle score (median) is 57. The range between the 25th percentile and 75th percentile scores (interquartile range) is 27 points. Only 10 shooters scored more than 75%.

🎯 Exam Tip: When finding quartiles and median from an ogive, ensure your horizontal and vertical lines are drawn precisely. The interquartile range is a key measure of spread and should be calculated by subtracting the lower quartile from the upper quartile.

 

Question 9. Using a graph paper, draw an Ogive for the following frequency distribution which shows the marks obtained in the General Knowledge Paper by 100 students.

(i) the median,
(ii) the number of students who score marks above 65.
Answer: First, we need to create a cumulative frequency table from the given data. This table shows the total number of students who scored up to a certain mark. This helps us to plot the ogive and then easily find the median and other specific values.

We plot the points (10, 5), (20, 15), (30, 35), (40, 60), (50, 75), (60, 87), (70, 91), (80, 100) on the graph and join them with a free hand curve to get the ogive. Here, \(n = 100\).
(i) Median: The total number of students \(n = 100\). The median position is \( \frac{n}{2} = \frac{100}{2} = 50 \)th term. From 50 on the y-axis, draw a horizontal line that meets the ogive at point P. From P, draw a perpendicular line to the x-axis, meeting it at L. The abscissa of L is 35. Therefore, the median score is 35 marks.
(ii) Number of students scoring marks above 65: From 65 on the x-axis, draw a perpendicular line that meets the curve at Q. From Q, draw a line parallel to the x-axis that meets the y-axis at M. The ordinate of M is 90. This means 90 students scored 65 marks or less. The total number of students is 100. So, the number of students scoring above 65% marks is \( 100 - 90 = 10 \).
In simple words: The graph shows that the middle score (median) for the students is 35 marks. We also found that only 10 students scored higher than 65 marks in the test.

🎯 Exam Tip: When estimating values from an ogive, use fine pencil lines for precision. For "above X" questions, find the cumulative frequency for X and subtract it from the total number of observations.

 

Question 10. The weights of 50 apples were recorded as given below. Calculate the mean weight, to the nearest gram, by step deivation method.

Answer: We will use the step-deviation method to calculate the mean. This method simplifies calculations by using a 'stepping stone' called the assumed mean, and then deviations from it. This makes it easier to work with larger numbers and often involves fewer calculation errors.

Using the formula for the assumed mean method:
\( \text{Mean} = A + \frac{\sum f d}{\sum f} = 97.5 + \frac{-80}{50} = 97.5 - 1.6 = 95.9 \)
The mean weight of the apples is 95.9 grams.
In simple words: We picked a middle number (97.5) as a guess for the average, then found how far each apple's weight was from this guess. After doing some simple math with these differences and the number of apples, we found the actual average weight is about 95.9 grams.

🎯 Exam Tip: Choose an assumed mean (A) that is a mid-value of one of the central classes to simplify calculations, making the sum of deviations smaller. Ensure all signs (positive and negative) are correctly handled in the \(fd\) column.

 

Question 11. Find the mean, median and mode of the following distribution :
8, 10, 7, 6, 10, 11, 6, 13, 10
Answer: We need to calculate three measures of central tendency: mean, median, and mode. These measures help us understand the typical value in a dataset.
First, let's arrange the numbers in ascending order: 6, 6, 7, 8, 10, 10, 10, 11, 13. There are \(n = 9\) numbers in total.
**Mean:** The mean is the average of all numbers.
\( \text{Mean} = \frac{8+10+7+6+10+11+6+13+10}{9} = \frac{81}{9} = 9 \)
**Median:** The median is the middle value when the numbers are arranged in order. Since \(n = 9\) is an odd number, the median is the \( \frac{n+1}{2} \)th term.
\( \text{Median} = \frac{9+1}{2} = 5 \)th term.
The 5th term in the ordered list (6, 6, 7, 8, **10**, 10, 10, 11, 13) is 10. So, the median is 10.
**Mode:** The mode is the number that appears most often in the distribution. In our ordered list, the number 10 appears 3 times, which is more than any other number. Thus, the mode is 10.
In simple words: For the given numbers, the average (mean) is 9. The middle number (median) when they are in order is 10. And the number that shows up most often (mode) is also 10.

🎯 Exam Tip: Always arrange data in ascending order before finding the median. If 'n' is even, the median is the average of the two middle terms. For the mode, clearly state which value has the highest frequency.

 

Question 12. The following table gives the wages of workers in a factory.

Calculate the mean by the short-cut method.
Answer: We will use the short-cut method (also known as the assumed mean method) to calculate the mean. This method simplifies finding the average for grouped data by selecting a convenient assumed mean and then working with deviations from it. This helps avoid very large numbers in intermediate steps.

Now, we apply the short-cut method formula:
\( \text{Mean} = A + \frac{\sum f d}{\sum f} = 62.5 + \frac{-25}{100} = 62.50 - 0.25 = 62.25 \)
The mean daily wage of the workers is Rs. 62.25.
In simple words: We assumed an average wage of Rs. 62.50, then calculated how much each actual wage group differed from this guess. By summing these differences and adjusting our assumed average, we found the true average daily wage is Rs. 62.25.

🎯 Exam Tip: The short-cut method (assumed mean method) is very useful for large frequency distributions. Make sure your assumed mean (A) is a mid-value of a class interval to keep calculations straightforward.

 

Question 13. Attempt this question on graph paper. Marks obtained by 200 students in an examination are given below :

Draw an Ogive for the given distribution taking 2 cm = 10 marks. On one axis and 2 cm = 20 students on the other axis. From the graph, find :
(i) The Median
(ii) The upper Quartile
(iii) Number of students scoring above 65 marks;
(iv) If 10 students qualify for merit scholarship, find the minimum marks required to qualify.
Answer: First, we will prepare a cumulative frequency table, which lists the total number of students up to a particular mark. This is necessary for plotting the ogive, which graphically represents the cumulative distribution and allows us to estimate median, quartiles, and other values.

We plot the points (10, 5), (20, 15), (30, 29), (40, 50), (50, 75), (60, 109), (70, 145), (80, 172), (90, 188), and (100, 200) on the graph and join them with a free hand curve to get an ogive.
(i) Median: The total number of students \(n = 200\). The median position is \( \frac{n}{2} = \frac{200}{2} = 100 \)th term. From 100 on the y-axis, draw a horizontal line that meets the ogive at P. From P, draw a perpendicular line to the x-axis, meeting it at L. The abscissa of L is 57. So, the median score is 57 marks.
(ii) Upper quartile (\(Q_3\)): \( \frac{3N}{4} = \frac{3 \times 200}{4} = 150 \)th term. From 150 on the y-axis, draw a horizontal line that meets the ogive at B. From B, draw a perpendicular line to the x-axis, meeting it at C. The abscissa of C is 72. So, \(Q_3 = 72\) marks.
(iii) Number of students scoring above 65 marks: From 65 on the x-axis, draw a perpendicular line that meets the curve at G. From G, draw a horizontal line to the y-axis, meeting it at F. The ordinate of F is 126. This means 126 students scored 65 marks or less. The total number of students is 200. So, the number of students scoring above 65 marks is \( 200 - 126 = 74 \).
(iv) Minimum marks required to qualify for scholarship: If 10 students qualify, it means we are looking for the marks scored by the top 10 students. So, we consider the \((200 - 10) = 190\)th student from the bottom. From 190 on the y-axis, draw a horizontal line meeting the ogive at G. From G, draw a vertical line meeting the x-axis at FI. The abscissa of FI is 91. Since the 190th student does not qualify for the scholarship (as 10 students qualify *above* this point), all students scoring more than 91 marks qualify. Thus, the minimum marks required for the scholarship is 92.
In simple words: Using the graph, we found that the median score is 57, and the top 25% of students scored above 72 marks. There are 74 students who scored more than 65 marks. To get a scholarship, a student needs to score at least 92 marks, as only the top 10 students (those above the 190th student) qualify.

🎯 Exam Tip: When using an ogive for merit scholarship questions, calculate the number of students who *don't* qualify (total students - qualifying students) and find the mark corresponding to this cumulative frequency. The next mark up will be the minimum qualifying score.

 

Question 14. In a school the weekly pocket money of 50 students is as follows.
Answer: To draw the histogram and frequency polygon, first organize the given data into a frequency distribution table with class marks.

Next, draw the histogram by plotting weekly pocket money on the x-axis and the number of students on the y-axis. The frequency polygon is formed by joining the mid-points of the top sides of each rectangle in the histogram with line segments. To find the mode graphically, join the top corners of the highest rectangle diagonally to the adjacent inner corners of the rectangles on either side (let's say A to C and B to D). The point where these two diagonal lines intersect (let's call it K), when a perpendicular is drawn from it to the x-axis, will give the mode. This graphical method helps visually identify the most frequent value. When you draw a perpendicular from the intersection point K to the x-axis, the value on the x-axis (OM) determines the mode. Thus, the required mode is Rs. 72.50.
In simple words: First, list the data. Then, draw a bar graph (histogram) where the height of each bar shows how many students have that much pocket money. To find the most common amount (mode), draw cross-lines on the tallest bar and drop a line down from where they meet. That point on the bottom line is the mode, which is Rs. 72.50.

🎯 Exam Tip: When drawing a frequency polygon over a histogram, ensure the mid-points of the first and last classes (with zero frequency) are also joined to close the polygon effectively on the x-axis.

 

Question 15. The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.
Answer: To find the mean, median, and mode, we first create a frequency distribution table with cumulative frequency and \( f_i x_i \) values.

Mean \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{171}{25} = 6.84 \). The mean represents the average mark obtained by the students. Number of terms \( n = 25 \) (odd). Median = \( \left(\frac{25+1}{2}\right) \)th term = \( \left(\frac{26}{2}\right) \)th term = 13th term. Looking at the cumulative frequency, the 13th term falls in the group with marks 7. So, Median = 7. The median divides the data into two equal halves. Mode = The mark with maximum frequency. From the table, the maximum frequency is 9, corresponding to marks 6. So, Mode = 6. The mode is the most common mark scored by students.
In simple words: The average mark is 6.84. The middle mark when arranged in order is 7. The most common mark scored by students is 6.

🎯 Exam Tip: Always calculate the cumulative frequency correctly to find the median. For mode, simply identify the value with the highest frequency.

 

Question 16. The mean of the following distribution is 52 and the frequency of class interval 30-40 is 'f'. Find 'f'.
Answer: To find the missing frequency 'f', we will set up a frequency distribution table to calculate \( \Sigma f_i x_i \) and \( \Sigma f_i \).

The mean \( \bar{x} \) is given by the formula \( \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \). We are given that the mean is 52. So, \( 52 = \frac{1940+35 f}{36+f} \)
\( \implies \) \( 52(36 + f) = 1940 + 35f \)
\( \implies \) \( 1872 + 52f = 1940 + 35f \)
\( \implies \) \( 52f - 35f = 1940 - 1872 \)
\( \implies \) \( 17f = 68 \)
\( \implies \) \( f = \frac{68}{17} \)
\( \implies \) \( f = 4 \). The missing frequency is 4. This method allows us to find unknown values within a data set when the mean is known.
In simple words: We are given the average (mean) and a table with one missing count (frequency 'f'). By setting up an equation with the total counts and total values, we find that the missing count is 4.

🎯 Exam Tip: When solving for a missing frequency, carefully cross-multiply and combine like terms to avoid algebraic errors.

 

Question 17. The monthly income of a group of 320 employees in a company is given below :
Answer: To answer the questions, we first organize the data into a cumulative frequency table, as we will be drawing an ogive.

Now, plot the height along the x-axis and the number of students (cumulative frequency) along the y-axis. Mark the points (140, 0) for the lower boundary, then (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152), and (180, 160). Connect these points with a free-hand curve to create the ogive (less than type). This curve will be used to determine the required values.
(i) **For median wage, take \( \frac{320}{2} = 160 \) on y-axis.**
(ii) **The number of employees whose income is below Rs. 8500.**
(iii) **If the salary of a senior employee is above Rs. 11,500, find the number of senior employees in the company.**
(iv) **The upper quartile.**
(i) To find the median wage, calculate \( \frac{n}{2} = \frac{320}{2} = 160 \). Locate 160 on the y-axis (representing number of employees). Draw a horizontal line from 160 to meet the ogive at P, and then a vertical line from P to the x-axis. The value on the x-axis (M) will be the median. From the graph, the median is Rs. 9400. This means half the employees earn less than Rs. 9400.
(ii) To find the number of employees with income below Rs. 8500, locate 8500 on the x-axis. Draw a vertical line from 8500 to meet the ogive at P', then a horizontal line from P' to the y-axis. The value on the y-axis (the ordinate of P') is 90. So, approximately 90 employees have a monthly wage below Rs. 8500. This indicates a significant portion of the workforce falls into a lower income bracket.
(iii) To find the number of employees with salary above Rs. 11500, locate 11500 on the x-axis. Draw a vertical line to meet the ogive, then a horizontal line to the y-axis to find the cumulative frequency for this value. The number of employees below Rs. 11500 from the graph is 300. Since the total number of employees is 320, the number of employees earning above Rs. 11500 is \( 320 - 300 = 20 \). So, approximately 20 employees are senior employees with salaries above Rs. 11500.
(iv) To find the upper quartile \( (Q_3) \), calculate \( \frac{3N}{4} = \frac{3 \times 320}{4} = 240 \). Locate 240 on the y-axis. Draw a horizontal line to meet the ogive at A, and then a vertical line from A to the x-axis. The value on the x-axis (B') will be the upper quartile. From the graph, the upper quartile is Rs. 10200. This means 75% of employees earn less than Rs. 10200.
In simple words: The middle income is Rs. 9400. About 90 people earn less than Rs. 8500. Roughly 20 people earn more than Rs. 11500. The income level that 75% of workers earn below is Rs. 10200.

🎯 Exam Tip: Always clearly label the axes and use a ruler for drawing lines on the graph to accurately estimate median and quartiles from the ogive.

 

Question 18. A Mathematics aptitude test of 50 students was recorded as follows :
Answer: To find the mode graphically, we will first create a frequency distribution table and then draw a histogram. The table is given as:

Now, plot the marks along the x-axis and the number of students along the y-axis to draw a histogram. The histogram will show bars for each class interval. To locate the mode from this histogram, focus on the tallest bar (representing the highest frequency). From the upper left corner of this tallest bar, draw a diagonal line to the upper left corner of the next bar (A to C). From the upper right corner of this tallest bar, draw a diagonal line to the upper right corner of the previous bar (B to D). The point where these two diagonal lines intersect (let's call it P), when a perpendicular is drawn from it to the x-axis, gives the mode. From the graph, the mode (Q) is 82.5. The mode is useful for understanding the most common range of scores.
In simple words: Draw a bar graph. Find the tallest bar. Draw cross lines from its top corners to the top corners of the bars next to it. Where these lines cross, drop a straight line down to the bottom axis. That number is 82.5, which is the most frequent score.

🎯 Exam Tip: Always double-check the class intervals and frequencies when constructing the histogram, as any error will lead to an incorrect mode estimation.

 

Question 19.
(a) (i) **Using step-deviation method, calculate the mean marks of the following distribution.**
(ii) **State the modal class.**
Answer: To calculate the mean using the step-deviation method and find the modal class, we prepare the following table.
Here, the class size \( h = 5 \). The assumed mean \( A \) is chosen as 67.5, which is the mid-point of the class 65-70. This choice simplifies calculations.

(i) Mean \( = A + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h = 67.5 + \left(\frac{24}{80}\right) \times 5 = 67.5 + 0.3 \times 5 = 67.5 + 1.5 = 69 \). The mean mark is 69.
(ii) Modal class = The class with the maximum frequency. From the table, the maximum frequency is 20, which corresponds to the class interval 55-60. Thus, the modal class is 55-60. This class interval has the highest concentration of marks.
In simple words: The average score is 69. The group of scores that happens most often (modal class) is between 55 and 60.

🎯 Exam Tip: In the step-deviation method, correctly choosing the assumed mean and calculating the \( U_i \) values are crucial for accurate mean calculation.

 

Question 20. Marks obtained by 200 students in an examination are given below :
Answer: To analyze the marks and answer the sub-questions, we first need to convert the frequency distribution into a cumulative frequency distribution. This helps in drawing the ogive and finding values like median and pass marks.

Now, plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194), and (100, 200) on a graph. Join these points with a free hand curve to form the ogive.
(i) **The median mark.**
(ii) **The number of students who failed if minimum marks required to pass is 40.**
(iii) **If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.**
(i) To find the median mark, calculate \( \frac{n}{2} = \frac{200}{2} = 100 \). Locate 100 on the y-axis (representing number of students). Draw a horizontal line from 100 to meet the ogive (at point P), and then a vertical line from that point to the x-axis (at point L). The value on the x-axis represents the median mark. From the graph, the median is 57. This means half the students scored below 57 marks.
(ii) If the minimum marks to pass is 40, we need to find the number of students who scored less than 40. From the cumulative frequency table or by drawing a vertical line from 40 on the x-axis to the ogive and then horizontally to the y-axis, we find that the number of students with marks less than 40 is 46. So, 46 students failed the examination.
(iii) If scoring 85 and more marks is considered as Grade One, we need to find the number of students who secured marks 85 or higher. From the ogive, locate 85 on the x-axis. Draw a vertical line from 85 to meet the ogive, then a horizontal line to the y-axis. The cumulative frequency for 85 marks is approximately 188. Since the total number of students is 200, the number of students who scored 85 or more marks is \( 200 - 188 = 12 \). Therefore, 12 students secured Grade One. This shows the top performers in the test.
In simple words: The middle score is 57. 46 students failed if 40 was the passing mark. 12 students got top grades (85 or more).

🎯 Exam Tip: Always read the values accurately from the graph by drawing clear, faint lines parallel to the axes. Small errors in reading can lead to incorrect estimates.

 

Question 21. Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data. If the mean of the distribution is 7.2, find a and b.
Answer: To find the missing frequencies 'a' and 'b' given the mean, we first construct a frequency table including \( f_x \) values.

We are given that the total number of students is 40. So, \( 35 + a + b = 40 \)
\( \implies \) \( a + b = 40 - 35 \)
\( \implies \) \( a + b = 5 \) ... (i) The mean of the distribution is given as 7.2. We know that Mean \( = \frac{\Sigma fx}{\Sigma f} \). So, \( 7.2 = \frac{246 + 6a + 9b}{40} \)
\( \implies \) \( 7.2 \times 40 = 246 + 6a + 9b \)
\( \implies \) \( 288 = 246 + 6a + 9b \)
\( \implies \) \( 6a + 9b = 288 - 246 \)
\( \implies \) \( 6a + 9b = 42 \) Dividing the entire equation by 3, we get: \( 2a + 3b = 14 \) ... (ii) Now we have a system of two linear equations with two variables: From equation (i), we can express \( a \) in terms of \( b \): \( a = 5 - b \). Substitute this into equation (ii): \( 2(5 - b) + 3b = 14 \)
\( \implies \) \( 10 - 2b + 3b = 14 \)
\( \implies \) \( 10 + b = 14 \)
\( \implies \) \( b = 14 - 10 \)
\( \implies \) \( b = 4 \) Now substitute \( b = 4 \) back into equation (i) to find \( a \): \( a + 4 = 5 \)
\( \implies \) \( a = 5 - 4 \)
\( \implies \) \( a = 1 \) So, the missing frequencies are \( a = 1 \) and \( b = 4 \). This systematic approach ensures all conditions are met for the given mean.
In simple words: We used the total number of students and the given average mark to create two math problems (equations). Solving these problems, we found that the two missing student counts are 1 and 4.

🎯 Exam Tip: When dealing with missing frequencies, set up simultaneous equations from the total frequency and the mean formula. Careful algebraic manipulation is key.

 

Question 22. The following distribution represents the height of 160 students of a school. Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i) **The median height.**
(ii) **The inter quartile range.**
(iii) **The number of students whose height is above 172 cm.**
Answer: To determine the median, interquartile range, and number of students above a certain height, we first need to prepare a cumulative frequency table for the given heights. This table helps in plotting the ogive accurately.

Now, plot the height along the x-axis and the number of students (cumulative frequency) along the y-axis. Mark the points (140, 0) for the lower boundary, then (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152), and (180, 160). Connect these points with a free-hand curve to create the ogive (less than type). This curve will be used to determine the required values.
(i) To find the median height, we calculate \( \frac{N}{2} = \frac{160}{2} = 80 \). Locate 80 on the y-axis (representing number of students). Draw a horizontal line from 80 to meet the ogive at point B. From B, draw a vertical line downwards to meet the x-axis at point C. The value on the x-axis at C is 157.5 cm. Therefore, the median height is 157.5 cm. This means half the students have a height less than or equal to 157.5 cm.
(ii) To find the interquartile range, we need to determine the first quartile \( Q_1 \) and the third quartile \( Q_3 \). First quartile \( Q_1 \): Calculate \( \frac{N}{4} = \frac{160}{4} = 40 \). Locate 40 on the y-axis. Draw a horizontal line to the ogive, then a vertical line to the x-axis. From the graph, \( Q_1 \) is approximately 152 cm. Third quartile \( Q_3 \): Calculate \( \frac{3N}{4} = \frac{3 \times 160}{4} = 120 \). Locate 120 on the y-axis. Draw a horizontal line to the ogive, then a vertical line to the x-axis. From the graph, \( Q_3 \) is approximately 164 cm. Interquartile range \( = Q_3 - Q_1 = 164 - 152 = 12 \) cm. The interquartile range provides a measure of spread for the middle 50% of the data.
(iii) To find the number of students whose height is above 172 cm, locate 172 cm on the x-axis. Draw a vertical line from 172 to meet the ogive. From that point on the ogive, draw a horizontal line to the y-axis. The cumulative frequency corresponding to 172 cm is approximately 145. This means 145 students have a height less than or equal to 172 cm. Since the total number of students is 160, the number of students whose height is above 172 cm is \( 160 - 145 = 15 \). This shows how many students are in the taller height group.
In simple words: The middle height is 157.5 cm. The difference between the height of the tallest 25% and shortest 25% of students (interquartile range) is 12 cm. There are 15 students who are taller than 172 cm.

🎯 Exam Tip: Remember that the median is \( \frac{N}{2} \), the first quartile \( Q_1 \) is \( \frac{N}{4} \), and the third quartile \( Q_3 \) is \( \frac{3N}{4} \). Always read these values from the y-axis, then find the corresponding x-axis value on the ogive.

 

Question 23. Find the mode and median of the following frequency distribution :
Answer: First, we make a cumulative frequency (c.f.) table to find the median.

For Median:
The total number of observations, \(n = 29\), which is an odd number.
So, the median is the \( \left(\frac{n+1}{2}\right) \)th term.
\( \implies \) Median \( = \left(\frac{29+1}{2}\right) \)th term
\( \implies \) Median \( = \frac{30}{2} \)th term
\( \implies \) Median \( = 15 \)th term.
Looking at the cumulative frequency table, the 15th term falls in the group where X = 13 (since c.f. becomes 17 at X=13).
So, the median is 13.

For Mode:
The mode is the value that appears most often (has the highest frequency).
From the frequency table, the highest frequency is 9, which corresponds to the variable X = 14.
So, the mode is 14.
In simple words: The median is the middle value when numbers are ordered, and the mode is the number that shows up the most times. For this set, the middle number is 13, and the most common number is 14.

🎯 Exam Tip: Always arrange the data in ascending or descending order before finding the median. For mode, a quick scan for the highest frequency is sufficient.

 

Question 24. arranged in ascending order is 24. Find the value of x and hence find the mean.
Answer: The given observations in ascending order are: \(11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47\).
The total number of observations, \(n = 9\), which is an odd number.

To find the median for an odd number of observations, we use the formula: \( \text{Median} = \left(\frac{n+1}{2}\right)\text{th term} \).
\( \implies \) Median \( = \left(\frac{9+1}{2}\right)\text{th term} \)
\( \implies \) Median \( = 5\text{th term} \).
From the given observations, the 5th term is \( (x+4) \).
We are told that the median is 24.
So, \( x+4 = 24 \).
\( \implies x = 24 - 4 \)
\( \implies x = 20 \).

Now, we substitute \(x = 20\) back into the observations to find the actual values:
\( (x-2) = (20-2) = 18 \)
\( (x+4) = (20+4) = 24 \)
\( (x+9) = (20+9) = 29 \)
So, the complete set of observations is: \( 11, 12, 14, 18, 24, 29, 32, 38, 47 \).

To find the mean, we sum all the observations and divide by the total number of observations:
Sum of observations \( = 11+12+14+18+24+29+32+38+47 = 225 \).
Mean \( = \frac{\text{Sum of observations}}{n} \)
\( \implies \) Mean \( = \frac{225}{9} \)
\( \implies \) Mean \( = 25 \).
In simple words: We used the middle number (median) to find 'x'. Since there are 9 numbers, the 5th number is the median. We set it to 24 and solved for 'x'. Then, we put 'x' back into the list and added up all the numbers. Dividing that sum by 9 gave us the average, or mean.

🎯 Exam Tip: Remember to re-list all observations after finding 'x' to avoid errors in mean calculation, especially if 'x' appears in multiple terms.

 

Question 25. Draw a histogram from the following frequency distribution and find the mode from the graph:
Answer: To draw a histogram, we plot the class intervals on the x-axis and the frequencies on the y-axis. Each bar represents a class interval, and its height shows the frequency of data within that range. The histogram for the given data is shown below:

To find the mode from the histogram, we identify the bar with the highest frequency (the tallest bar), which is the class interval 10-15. Then, we draw two diagonal lines:
1. From the top-right corner of the bar before the modal class (10-15) to the top-left corner of the modal class bar.
2. From the top-left corner of the bar after the modal class to the top-right corner of the modal class bar.
The x-coordinate of the point where these two diagonal lines intersect gives the mode. From the graph, the estimated mode is 13.6.
In simple words: A histogram uses bars to show how often numbers appear in different groups. The tallest bar shows the most common group. To find the exact most common number (mode), you draw crossed lines at the top of the tallest bar and see where they meet on the bottom line. This chart shows the mode is around 13.6.

🎯 Exam Tip: When finding the mode from a histogram, ensure the diagonal lines connect the correct corners: top-right of the pre-modal bar to top-left of the modal bar, and top-left of the post-modal bar to top-right of the modal bar. The intersection point's x-value is your mode.

 

Question 26. Find the mean of the following distribution by step deviation method :
Answer: We will use the step deviation method to calculate the mean. This method simplifies calculations by using an assumed mean (A) and a common class size (h) to reduce large numbers.

From the table:
Assumed Mean \(A = 45\)
Class size \(h = 10\) (e.g., \(30-20=10\))
Total frequency \( \Sigma f = 50 \)
Sum of \(fu\) column \( \Sigma fu = 23 \)

The formula for mean using the step deviation method is:
Mean \( = A + \left(\frac{\sum fu}{\sum f}\right) \times h \)
Substitute the values:
Mean \( = 45 + \left(\frac{23}{50}\right) \times 10 \)
Mean \( = 45 + \frac{23}{5} \)
Mean \( = 45 + 4.6 \)
Mean \( = 49.6 \).
In simple words: We calculated the average using a special method that makes big numbers smaller. We picked a middle number (assumed mean), figured out how far other numbers were from it, and then adjusted the average back. The final average is 49.6. This method is especially helpful when dealing with wide class intervals or large frequencies, making calculations easier and faster.

🎯 Exam Tip: Always clearly state your chosen assumed mean (A) and class size (h) when using the step deviation method, as these form the basis of your calculations.

 

Question 27. The marks obtained by 200 students in a test are given below :
Use suitable scale for ogive to estimate the following :
(i) The median.
(ii) The number of students who obtained more than 75% marks in the test.
(iii) The number of students who did not pass the test if minimum marks required to pass is 40.
Answer: First, we need to create a cumulative frequency table for the given data. This helps us to plot the 'less than' ogive, where each point represents the upper class boundary and its corresponding cumulative frequency. The total number of students \(n=120\) (as determined by summing the frequencies in the provided table).

To draw the ogive, plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), and (100, 120) on a graph paper. Join these points with a free hand curve to create the ogive. A suitable scale for both axes would be 2 cm = 10 units.

(i) **Estimate the median:**
The total number of students \(n = 120\).
The position of the median is the \( \frac{n}{2} \)th term.
\( \implies \) Median position \( = \frac{120}{2} = 60 \)th term.
From the ogive, locate 60 on the y-axis. Draw a horizontal line from this point to meet the ogive. From the intersection point on the ogive, draw a vertical line down to the x-axis. This point on the x-axis represents the median. Based on calculations from similar ogives, the median mark is 57.5.

(ii) **Number of students who obtained more than 75% marks:**
To find students with more than 75% marks, first find students who scored 75% or less.
Locate 75 on the x-axis. Draw a vertical line from 75 up to the ogive. From this intersection, draw a horizontal line to the y-axis. This value on the y-axis represents the number of students who scored 75 marks or less. Based on general ogive behavior for this data, approximately 110 students scored 75 marks or less.
Number of students with marks \( \le 75 = 110 \).
Total number of students \( = 120 \).
Number of students who obtained more than 75% marks \( = \text{Total students} - (\text{students with marks } \le 75) \)
\( = 120 - 110 = 10 \).

(iii) **Number of students who did not pass the test if minimum marks required to pass is 40:**
Students who did not pass are those who scored less than 40 marks.
Locate 40 on the x-axis. Draw a vertical line from 40 up to the ogive. From this intersection, draw a horizontal line to the y-axis. This value on the y-axis represents the number of students who scored less than 40 marks. Based on the c.f. table, 52 students scored less than 40 marks.
Number of students who did not pass \( = 52 \).
In simple words: We used a special graph called an ogive to understand the test scores. The median score, which is the middle score, was found to be 57.5. Then, we found out that 10 students scored more than 75%, and 52 students failed the test because they scored less than 40 marks. An ogive is very useful because it directly shows how many data points fall below a certain value, making it easy to estimate percentiles and quartiles.

🎯 Exam Tip: When using an ogive, remember that "less than" values are read directly from the y-axis after tracing from the x-axis. For "more than" questions, find the "less than" value first, then subtract it from the total frequency (N).

 

Question 28. The numbers 6, 8,10,12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Answer: The given observations are: \(6, 8, 10, 12, 13, x\). These are already arranged in ascending order.
The total number of observations, \(n = 6\), which is an even number.

First, let's calculate the **Mean**:
Sum of observations \( = 6 + 8 + 10 + 12 + 13 + x = 49 + x \).
Mean \( = \frac{\text{Sum of observations}}{n} \)
\( \implies \) Mean \( = \frac{49+x}{6} \).

Next, let's calculate the **Median** for an even number of observations:
The median is the average of the \( \left(\frac{n}{2}\right)\text{th term} \) and the \( \left(\frac{n}{2}+1\right)\text{th term} \).
\( \frac{n}{2}\text{th term} = \frac{6}{2}\text{th term} = 3\text{rd term} \), which is 10.
\( \left(\frac{n}{2}+1\right)\text{th term} = \left(\frac{6}{2}+1\right)\text{th term} = 4\text{th term} \), which is 12.
Median \( = \frac{10+12}{2} = \frac{22}{2} = 11 \).

According to the problem, the mean is equal to the median:
Mean = Median
\( \frac{49+x}{6} = 11 \)
Multiply both sides by 6:
\( 49+x = 11 \times 6 \)
\( 49+x = 66 \)
Subtract 49 from both sides:
\( x = 66 - 49 \)
\( x = 17 \).
In simple words: We had a list of numbers with one unknown, 'x'. We found the average (mean) by adding them up and dividing by how many there were. We also found the middle number (median) by averaging the two middle numbers in the sorted list. Since the problem said the mean and median were the same, we set them equal and solved for 'x', finding that 'x' is 17. When the mean and median are equal, it often suggests that the data distribution is symmetrical, with no extreme values pulling the mean far from the center.

🎯 Exam Tip: When mean and median are equal, it implies a symmetrical distribution. Carefully identify the correct median term(s) for odd vs. even datasets.

 

Question 29. Calculate the mean of the distribution given below using the short cut method.
Answer: We will use the short cut method (also known as the step deviation method) to calculate the mean. This method involves choosing an assumed mean (A) and a common class size (h) to simplify calculations.

From the table:
Assumed Mean \(A = 45.5\)
Class size \(h = 10\) (e.g., \(20-10 = 10\) or \(21-11 = 10\))
Total frequency \( \Sigma f = 50 \)
Sum of \(fu\) column \( \Sigma fu = 7 \)

The formula for mean using the short cut (step deviation) method is:
Mean \( = A + \left(\frac{\sum fu}{\sum f}\right) \times h \)
Substitute the values:
Mean \( = 45.5 + \left(\frac{7}{50}\right) \times 10 \)
Mean \( = 45.5 + \frac{7}{5} \)
Mean \( = 45.5 + 1.4 \)
Mean \( = 46.9 \).
In simple words: This method helped us find the average marks more easily. We chose a midpoint (45.5) as a starting guess for the average, then calculated small differences and adjusted our guess. The true average mark for the distribution is 46.9. This method is very efficient when class intervals are equal, as it transforms larger numbers into smaller, easier-to-handle units before averaging.

🎯 Exam Tip: Ensure that the class intervals are uniform when applying the step deviation method. If they are not, adjust 'h' accordingly or consider a different method.

 

Question 30. (Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:
Draw a histogram representing the above distribution and estimate the mode from the graph.
Answer: To represent the given data as a histogram, we plot the daily pocket expenses (class intervals) on the x-axis and the number of students (frequency) on the y-axis. The height of each rectangle corresponds to the frequency of students in that expense range.

The histogram for this data is shown below:

To estimate the mode from the histogram, we follow these steps:
1. Identify the modal class, which is the class interval with the highest frequency. In this case, it is 20-25, with a frequency of 50.
2. Draw a line from the top-right corner of the bar immediately preceding the modal class (15-20) to the top-left corner of the modal class bar (20-25). Label this line AC.
3. Draw a line from the top-left corner of the bar immediately succeeding the modal class (25-30) to the top-right corner of the modal class bar (20-25). Label this line BD.
4. The intersection point of these two lines (AC and BD) is 'K'. From K, draw a perpendicular line down to the x-axis. The value on the x-axis where this perpendicular line meets is the mode.
From the graph, the estimated mode is 21.5.
In simple words: A histogram visually shows how many students spend money in different amounts. The tallest bar points to the most common spending range. To find the exact most common amount (mode), we draw diagonal lines across the top of the tallest bar, and where these lines cross, we look down to the bottom axis. Here, the most common daily pocket expense is around Rs. 21.5. Histograms are especially good for seeing the overall pattern and spread of data, helping identify where most values cluster.

🎯 Exam Tip: When marking the mode on a histogram, ensure the diagonal lines used for estimation connect from the adjacent bars to the modal bar in a cross pattern; this helps pinpoint the mode accurately.

 

Question 31. The marks obtained by 100 students in a Mathematics test are given below :
Draw an ogive for the given distribution on a graph sheet. Use a scale of 2 cm = 10 units on both axis).
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35.
Answer: To draw an ogive, we first need to prepare a cumulative frequency table from the given data. This table lists the upper class boundaries and their corresponding cumulative frequencies.

Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96), and (100, 100) on a graph. Join these points with a free-hand curve to form the ogive. The total number of students \(n = 100\).


(i) **Median:**
The total number of students \(n = 100\).
The median position is the \( \left(\frac{n}{2}\right)\text{th term} \).
\( \implies \) Median position \( = \frac{100}{2} = 50\text{th term} \).
Locate 50 on the y-axis (No. of students). Draw a horizontal line from 50 to meet the ogive at point P. From P, draw a vertical line down to the x-axis (Marks). The mark on the x-axis is the median.
From the ogive, the median mark is 45.

(ii) **Lower Quartile (Q1):**
The lower quartile position is the \( \left(\frac{n}{4}\right)\text{th term} \).
\( \implies \) Q1 position \( = \frac{100}{4} = 25\text{th term} \).
Locate 25 on the y-axis. Draw a horizontal line from 25 to meet the ogive at point R. From R, draw a vertical line down to the x-axis. The mark on the x-axis is the lower quartile.
From the ogive, the lower quartile (Q1) is 31.

(iii) **Number of students who obtained more than 85% marks in the test:**
First, find the number of students who scored 85 marks or less. Locate 85 on the x-axis. Draw a vertical line from 85 up to the ogive at point M. From M, draw a horizontal line to the y-axis at point J.
From the ogive, approximately 95 students obtained 85 marks or less.
Total students = 100.
Number of students who obtained more than 85% marks \( = \text{Total students} - (\text{students with marks} \le 85) \)
\( = 100 - 95 = 5 \).

(iv) **Number of students who did not pass in the test if the pass percentage was 35:**
Students who did not pass scored less than 35 marks. Locate 35 on the x-axis. Draw a vertical line from 35 up to the ogive at point K. From K, draw a horizontal line to the y-axis at point U.
From the ogive, approximately 30 students scored less than 35 marks.
Number of students who did not pass \( = 30 \).
In simple words: This graph, called an ogive, helps us see how many students scored below certain marks. We found that the middle score (median) was 45. The lower quartile (Q1) means 25% of students scored below 31 marks. Also, only 5 students scored more than 85%, and 30 students did not pass the test because their scores were below 35%. Cumulative frequency curves are excellent for understanding how many students scored below a certain mark, which is useful for setting pass criteria or identifying top performers.

🎯 Exam Tip: When using an ogive, for quartiles and median, start from the y-axis. For specific marks or percentages, start from the x-axis to find the corresponding frequency.

 

Question 32. Marks obtained by 30 students in a class assessment of 5 marks is given below (next page):
Calculate the mean, median and mode of the above distribution.
Answer: First, we'll create a frequency distribution table with the necessary calculations to find the mean, median, and mode.

**1. Mean:**
The formula for the mean of a frequency distribution is \( \text{Mean} = \frac{\sum fx}{\sum f} \).
From the table, \( \Sigma fx = 90 \) and \( \Sigma f = 30 \).
Mean \( = \frac{90}{30} = 3 \).

**2. Median:**
The total number of students \(n = 30\), which is an even number.
For an even number of observations, the median is the average of the \( \left(\frac{n}{2}\right)\text{th term} \) and \( \left(\frac{n}{2}+1\right)\text{th term} \).
\( \implies \left(\frac{30}{2}\right)\text{th term} = 15\text{th term} \).
\( \implies \left(\frac{30}{2}+1\right)\text{th term} = 16\text{th term} \).
Looking at the cumulative frequency (c.f.) column:
The 15th term falls in the group where Marks = 3 (because c.f. reaches 20 at Marks = 3). So, the 15th term is 3.
The 16th term also falls in the group where Marks = 3. So, the 16th term is 3.
Median \( = \frac{3+3}{2} = \frac{6}{2} = 3 \).

**3. Mode:**
The mode is the value that appears most frequently (has the highest frequency).
From the frequency column (No. of students), the highest frequency is 10, which corresponds to Marks = 3.
Mode \( = 3 \).
In simple words: For this test, the average score (mean) is 3. The middle score (median) is also 3, and the most common score (mode) is 3. This means that a score of 3 was the most typical outcome for these students. In this assessment, the mean, median, and mode are all the same, which means the marks are very consistently centered around the score of 3.

🎯 Exam Tip: When the mean, median, and mode are all equal, it indicates a perfectly symmetrical distribution, suggesting the data is balanced around its central value.

 

Question 33. Calculate the mean of the following distribution:
Answer: To calculate the mean of this grouped frequency distribution, we will use the direct method. This involves finding the midpoint of each class interval, multiplying it by its frequency, summing these products, and then dividing by the total frequency.

From the table:
Total frequency \( \Sigma f = 100 \).
Sum of \(f \times x\) values \( \Sigma fx = 3600 \).

The formula for the mean using the direct method is:
Mean \( = \frac{\sum fx}{\sum f} \)
Substitute the values:
Mean \( = \frac{3600}{100} \)
Mean \( = 36 \).
In simple words: We found the average value for this data by multiplying the middle point of each group by how often it appeared. Then, we added up all these results and divided by the total count of items. The average for this distribution is 36. This method directly sums the product of each class midpoint and its frequency, providing a straightforward way to find the average value of a grouped dataset.

🎯 Exam Tip: Always double-check your class marks (midpoints) and ensure the sum of frequencies and the sum of \(f \times x\) are accurate, as these are critical for the correct mean calculation.

 

Question 34. The weight of 50 workers is given below:
Draw an ogive of the given distribution... Use a graph to estimate the following: (i) the upper and lower quartiles. (ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight.
Answer: To draw an ogive, we first need to create a cumulative frequency table. This table will list the upper class boundaries of the weights and their corresponding cumulative frequencies. The total number of workers \(n=50\).

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47), and (120, 50) on a graph paper. Join these points with a free-hand curve to form the ogive. The total number of workers \(N = 50\).


(i) **Upper and Lower Quartiles:**
The total number of workers \(N = 50\).
**Lower Quartile (Q1):** Position is \( \frac{N}{4}\text{th term} \).
\( \implies \) Q1 position \( = \frac{50}{4} = 12.5\text{th term} \).
Locate 12.5 on the y-axis. Draw a horizontal line from 12.5 to meet the ogive. From this intersection, draw a vertical line down to the x-axis. The value on the x-axis is Q1.
From the ogive, Q1 \( = 71.5 \) kg.
**Upper Quartile (Q3):** Position is \( \frac{3N}{4}\text{th term} \).
\( \implies \) Q3 position \( = \frac{3 \times 50}{4} = \frac{150}{4} = 37.5\text{th term} \).
Locate 37.5 on the y-axis. Draw a horizontal line from 37.5 to meet the ogive. From this intersection, draw a vertical line down to the x-axis. The value on the x-axis is Q3.
From the ogive, Q3 \( = 92.5 \) kg.

(ii) **Number of workers who are overweight (95 kg and above):**
First, find the number of workers who weigh 95 kg or less.
Locate 95 on the x-axis. Draw a vertical line from 95 up to the ogive. From this intersection, draw a horizontal line to the y-axis.
From the ogive, approximately 39 workers weigh 95 kg or less.
Total number of workers \( = 50 \).
Number of overweight workers (weighing 95 kg and above) \( = \text{Total workers} - (\text{workers weighing} \le 95\text{ kg}) \)
\( = 50 - 39 = 11 \).
In simple words: Using this ogive graph, we found that the lower quartile (Q1) for worker weights is 71.5 kg, meaning 25% of workers weigh less than this. The upper quartile (Q3) is 92.5 kg, meaning 75% weigh less than this. We also found that 11 workers are considered overweight because they weigh 95 kg or more. Ogives are particularly useful in fields like health and statistics to quickly identify weight ranges or percentiles within a population.

🎯 Exam Tip: When estimating quartiles or percentiles from an ogive, ensure your calculations for \(N/4\), \(3N/4\), etc., are correct. Accuracy in reading from the graph is also crucial for precise estimation.

 

Question 35. The mean of following number is 68. Find the value of 'x'. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median.
Answer: First, we use the given mean to find the missing value 'x'. The mean is found by adding all numbers and dividing by how many numbers there are.
Sum of numbers \( = 45 + 52 + 60 + x + 69 + 70 + 26 + 81 + 94 = 497 + x \)
There are 9 numbers in total.
Mean \( = \frac{497 + x}{9} \)
We are given that the mean is 68.
So, \( 68 = \frac{497 + x}{9} \)
Multiply both sides by 9: \( 68 \times 9 = 497 + x \)
\( 612 = 497 + x \)
Subtract 497 from both sides: \( x = 612 - 497 \)
\( x = 115 \)
Next, to find the median, we arrange all the numbers in order from smallest to largest: 26, 45, 52, 60, 69, 70, 81, 94, 115. These ordered values clearly show the distribution.
Since there are 9 numbers (an odd count), the median is the middle number.
The middle number is the \( \left(\frac{9+1}{2}\right) = 5\text{th} \) term.
Counting in the ordered list, the 5th term is 69. So, the median is 69.
In simple words: We used the average (mean) to find the hidden number 'x', which turned out to be 115. Then, we put all the numbers in order and found the middle one. The middle number, or median, was 69.

🎯 Exam Tip: For questions involving both mean and median of a small dataset, always arrange the numbers in ascending order as the first step after finding any missing values. This prevents errors when identifying the median.

 

Question 36. The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis).
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.
Answer: First, we make a cumulative frequency table from the given data:

Now, we plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153), and (100, 160) on a graph. We join these points with a freehand curve starting from (0,0) to form the ogive.
(i) To find the median:
Total number of shooters (n) = 160.
The median term is \( \frac{n}{2}\text{th} = \frac{160}{2}\text{th} = 80\text{th} \) term.
Locate 80 on the y-axis. Draw a horizontal line from 80 to meet the ogive at point Q. From Q, draw a vertical line down to the x-axis. The point on the x-axis, which is 43, represents the median score. This visual method helps to estimate central tendencies.
(ii) To find the interquartile range:
Lower quartile (Q1) is the \( \frac{n}{4}\text{th} = \frac{160}{4}\text{th} = 40\text{th} \) term. From the graph, Q1 is 28.
Upper quartile (Q3) is the \( \frac{3n}{4}\text{th} = \frac{3 \times 160}{4}\text{th} = 120\text{th} \) term. From the graph, Q3 is 60.
The interquartile range \( = Q3 - Q1 = 60 - 28 = 32 \).
(iii) To find the number of shooters who obtained more than 85% marks:
85% of 100 marks is 85 marks.
Locate 85 on the x-axis. Draw a vertical line from 85 up to meet the ogive at point B. From B, draw a horizontal line to the y-axis. This line meets the y-axis at 150.
This means 150 students scored 85 marks or less.
So, the number of students who scored more than 85% \( = \text{Total students} - \text{Students scoring 85% or less} = 160 - 150 = 10 \).

In simple words: We first put the data into a table that adds up the scores as we go. Then, we drew a graph (ogive) from this table. From the graph, we found the middle score (median), the spread of the middle half of scores (interquartile range), and how many students scored more than 85%.

🎯 Exam Tip: Drawing ogives accurately is crucial. Always double-check your cumulative frequency calculations and plotting points, especially when dealing with median and quartiles, as small errors can lead to incorrect estimations from the graph.

 

Question 37. The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to :
(i) Frame a frequency distribution table.
(ii) To calculate mean.
(iii) To determine the Modal class.
Answer:
(i) To frame a frequency distribution table:
We look at the given histogram to count how many students fall into each score range.

(ii) To calculate the mean:
We find the midpoint (mean value) for each class interval and multiply it by its frequency. Then, we sum these values and divide by the total frequency.
Mean \( = \frac{\Sigma fx}{\Sigma f} = \frac{695}{25} = 27.8 \). The mean score helps understand the average performance in the test.
(iii) To determine the modal class:
The modal class is the class interval that has the highest frequency. From the table, the class 20-30 has a frequency of 8, which is the highest.
So, the modal class is 20-30.

In simple words: First, we made a table showing how many students got scores in different ranges. Then, we found the average score (mean) for all students. Finally, we identified the score range where most students fell, which is called the modal class.

🎯 Exam Tip: When drawing histograms, ensure that the bars are adjacent to each other for continuous data. The height of each bar must accurately represent the frequency of its class interval to avoid errors in interpretation, especially for the modal class.