CBSE Class 10 Mathematics Statistics Worksheet Set 05

Question. To represent ‘the less than type’ graphically, we plot the ________ on the x – axis.
(a) class marks
(b) class size
(c) lower limits
(d) upper limits
Answer: (d) upper limits
Explanation: To represent ‘the less than type’ graphically, we plot the upper limits on the x-axis.
e.g marks obtained by students are represented in grouped data as (0 - 10), (10 - 20), (20 - 30), (30 - 40) .........
only upper limits such as 10, 20, 30, 40 ....... are taken for the x-axis

Question. \( \frac{Upper class limit + Lower class limit}{2} = \).
(a) frequency
(b) Class mark
(c) None of the options
(d) class size
Answer: (b) Class mark
Explanation: In each class interval of grouped data, there are two limits or boundaries (upper limit and lower limit) while the mid-value is equal to \( \frac{Upper class limit + Lower class limit}{2} \). These mid-values are also known as Classmark.

Question. The mean of the first 10 prime numbers is
(a) 129
(b) 1.29
(c) 12.9
(d) 11.9
Answer: (c) 12.9
Explanation: The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
\( \therefore \text{ Mean} = \frac{\text{ Sum of first 10 prime numbers}}{10} \)
\( = \frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29}{10} = \frac{129}{10} = 12.9 \)

Question. The mean of the first 10 natural odd numbers is
(a) 9
(b) 12
(c) 11
(d) 10
Answer: (d) 10
Explanation: The first 10 natural odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
\( \therefore \text{ Mean} = \frac{\text{ Sum of first 10 natural odd numbers}}{10} \)
\( = \frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10} = \frac{100}{10} = 10 \)

Question. For the following distribution

the sum of lower limits of the median class and modal class is
(a) 190
(b) 20
(c) 180
(d) 170
Answer: (d) 170
Explanation:

Here \( N = 66 \Rightarrow \frac{N}{2} = 33 \). The median class is 80 – 90 and Modal class is 90 – 100.
Sum of lower limits of Median class and Modal class = 80 + 90 = 170

Question. If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, find the median.
Answer: Mean = 53.4, Mode = 55.2
Mode = 3 Median - 2 Mean
Hence, \( \text{ Median} = \frac{\text{Mode} + 2 \text{ Mean}}{3} = \frac{55.2 + 2(53.4)}{3} = \frac{55.2 + 106.8}{3} = \frac{162}{3} = 54 \)

Question. Find the mode of the following data:
15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Answer: Mode : It is the value which occurs maximum number of times.
Given data: 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15

From above table, Mode = 15 because it occurs maximum number of times.

Question. In the table given below, the times taken by 120 athletes to run a 100-m hurdle race are given.

Find the number of athletes who completed the race in less than 14.6 seconds.
Answer: According to the question,
Number of athletes who completed the race in less than 14.6 seconds are
\( = 2 + 4 + 15 + 54 = 75 \)
Therefore, 75 athletes completed the race in less than 14.6 seconds.

Question. What is the lower limit of the modal class of the following frequency distribution?

Answer: The Class having maximum frequency is called as modal class.
From observing table, Here, maximum frequency = 27.
\( \therefore \) The modal class is 40 - 50.
The lower limit of the modal class is 40.

Question. Find the class marks of classes 10 - 25 and 35 - 55.
Answer: Class - mark of class 10 - 25 = \( \frac{10 + 25}{2} = \frac{35}{2} = 17.5 \)
Class - mark of class 35 - 55 = \( \frac{35 + 55}{2} = \frac{90}{2} = 45 \)

Question. Convert the following cumulative distribution to a frequency distribution:

Answer:

Question. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Answer:

Using the direct method,
\( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{499}{40} = 12.475 \)
Hence, the mean number of days a student was absent is 12.48.

Question. Candidates of four schools appear in a mathematics test. The data were as follow:

If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Answer: Let the number of candidates from school III = P

Given Average score for all schools = 66
\( \frac{N_1\bar{x}_1 + N_2\bar{x}_2 + N_3\bar{x}_3 + N_4\bar{x}_4}{N_1 + N_2 + N_3 + N_4} = 66 \)
\( \frac{60(75) + 48(80) + P(55) + 40(50)}{60 + 48 + P + 40} = 66 \)
\( \frac{4500 + 3840 + 55p + 2000}{148 + P} = 66 \)
\( \Rightarrow 4500 + 3840 + 55p + 2000 = 66(60 + 48 + p + 40) \)
\( \Rightarrow 10340 + 55p = 66p + 9768 \)
\( \Rightarrow 10340 - 9768 = (66 - 55)p \)
\( \Rightarrow 11P = 572 \)
\( \Rightarrow P = 52 \)

Question. Write the proper data (marks) and frequency (Number of students) for the following cumulative frequency distribution:

Answer: Table:

Question. During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarised as follows:

Find the mean of heartbeats per minute for these patients, choosing a suitable method.
Answer: Following table shows the given data & assumed mean deviation method to calculate the mean :-

Let, assumed mean ( A ) = 75.5.....(1)
Now, from table :- \( \sum f_i = 30 \) and \( \sum f_i d_i = 12 \).....(2)
Now, \( \text{mean} = A + \frac{\sum f_i d_i}{\sum f_i} \)
\( = 75.5 + \frac{12}{30} \)
\( = 75.5 + 0.4 = 75.9 \)
Thus, the mean of heartbeats per minute for these patients is 75.9

Question. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Find the average expenditure (in Rs.) per household.
Answer:

Let the assumed mean be A=325.
N = 200, A = 325, h = 50, and \( \sum f_i u_i = -235 \)
\( mean = \bar{x} = A + h \frac{1}{N} \sum f_i u_i \)
\( \Rightarrow \bar{x} = 325 + 50 \times \left\{ \frac{-235}{200} \right\} \)
\( \Rightarrow \bar{x} = 325 - \frac{235}{4} = 325 - 58.75 = 266.25 \)

Question. The percentage of various categories of workers in a state is given in the following table. Present the information in the form of a pie chart

Answer:

Question. Draw 'more than' ogive for the following distribution. Find the median from the curve:

Answer:

From graph, \( N = 100, \frac{N}{2} = 50 \). Median = 25.

Question. Find the mode, median and mean for the following data:

Answer:

i. Mean = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{4970}{100} = 49.70 \)
ii. Median: N = 100, \( \frac{N}{2} = 50 \). Median Class is 45 - 55.
\( \text{Median} = l + \left( \frac{\frac{N}{2} - c}{f} \right) \times h = 45 + \left\{ 10 \times \frac{50 - 38}{33} \right\} = 45 + 3.64 = 48.64 \)
iii. we know that, \( \text{Mode} = 3 \times \text{median} - 2 \times \text{mean} \)
\( = 3 \times 48.64 - 2 \times 49.70 = 145.92 - 99.4 = 46.52 \)

Question. The median of the following data is 525. Find the values of x and y if the total frequency is 100.

Answer: Table:

N = 100, Hence, \( 76 + x + y = 100 \Rightarrow x + y = 24 \) .....(i)
Given, Median = 525, which lies between class 500 - 600
\( \Rightarrow \text{Median class} = 500 - 600 \)
Now, \( \text{Median} = l + \frac{\frac{n}{2} - c.f}{f} \times h \)
\( 525 = 500 + \left[ \frac{50 - (36 + x)}{20} \right] \times 100 \)
\( \Rightarrow 25 = (50 - 36 - x) \times 5 \)
\( \Rightarrow 5 = 14 - x \)
\( \Rightarrow x = 9 \)
Substituting the value of x in equation (i),
\( y = 24 - 9 = 15 \)
Hence, x = 9 and y = 15

Question. The mean of ‘n’ observations is \( \bar{x} \). If the first item is increased by 1, second by 2 and so on, then the new mean is
(a) \( \bar{x} - \frac{n-1}{2} \)
(b) \( \bar{x} - \frac{n+1}{2} \)
(c) \( \bar{x} + \frac{n+1}{2} \)
(d) \( \bar{x} \)
Answer: (c) \( \bar{x} + \frac{n+1}{2} \)
Explanation: Let terms be \( x_1, x_2, x_3, ............., x_n \).
\( \therefore \text{Mean } (\bar{x}) = \frac{x_1 + x_2 + x_3 + ............ + x_n}{n} \)
\( \Rightarrow x_1 + x_2 + x_3 + ............ + x_n = n.\bar{x} \)
New observations are \( x_1 + 1, x_2 + 2, x_3 + 3, ............., x_n + n \).
\( \therefore \text{New Mean} = \frac{(x_1+1)+(x_2+2)+(x_3+3)+...........+(x_n+n)}{n} \)
\( = \frac{(x_1+x_2+x_3+...........+x_n) + (1+2+3+.......+n)}{n} \)
\( = \frac{n.\bar{x} + \frac{n(n+1)}{2}}{n} = \bar{x} + \frac{n+1}{2} \)

Question. The arithmetic mean of a set of 40 values is 65. If each of the 40 values is increased by 5, what will be the mean of the set of new values:
(a) 65
(b) 60
(c) 70
(d) 50
Answer: (c) 70
Explanation: Mean of 40 values = 65
Total of 40 values = \( 65 \times 40 = 2600 \)
When each value is increased by 5, then total of 40 values will be \( 40 \times 5 = 200 \) more than 2600
\( \therefore \text{New total of 40 values} = 2600 + 200 = 2800 \)
Now, \( \text{New mean of 40 values} = \frac{2800}{40} = 70 \)

Question. The most frequent value in the data is known as
(a) mean
(b) mode
(c) All of the options
(d) median
Answer: (b) mode
Explanation: The most frequent value in the data is known as the Mode.
e.g let us consider the following data set: 3,5,7,5,9,5,8,4
the mode is 5, since it occurs most often in data set.

Question. If \( x_i \)'s are the midpoints of the class intervals of grouped data, \( f_i \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_i x_i - \bar{x}) \) is equal to
(a) 2
(b) 0
(c) -1
(d) 1
Answer: (b) 0
Explanation: If \( x_i \)'s are the midpoints of the class intervals of grouped data, \( f_i \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( \sum (f_i x_i - \bar{x}) \) is equal to 0. i.e the difference between the sum of product of frequencies and mid values of corresponding class intervals of the grouped data and the sum of their mean value is equal to zero.

Question. The marks obtained by 9 students in Mathematics are 59, 46, 30, 23, 27, 40, 52, 35 and 29. The median of the data is
(a) 29
(b) 35
(c) 40
(d) 30
Answer: (b) 35
Explanation: Arranging the given data in ascending order, we get
23, 27, 29, 30, 35, 40, 46, 52, 59
Here, \( n = 9 \), which is odd.
\( \therefore \text{Median} = \left( \frac{n+1}{2} \right)^{\text{th}} \text{ term} \)
\( = \left( \frac{9+1}{2} \right)^{\text{th}} \text{ term} = 5^{\text{th}} \text{ term} = 35 \)

Question. For a particular year, the following is the distribution of the age (in yrs.) of primary school teachers in H.P:

Find how many teachers are of age less than 31 years.
Answer: Number of teachers of age less than 31 years \( = 11 + 32 + 51 = 94 \)

Question. The widths of 50 leaves of a plant were measured in mm and their cumulative frequency distribution is shown in the following table. Make frequency distribution table for this.

Answer: Table:

Question. A data has 25 observations arranged in a descending order. Which observation represents the median?
Answer: Number of observations \( = 25 \)
Hence, median \( = \left( \frac{n+1}{2} \right)^{\text{th}} \text{ Observation} \)
\( = \left( \frac{25+1}{2} \right)^{\text{th}} \text{ Observations} = \text{Value of } 13^{\text{th}} \text{ Observation} \)

Question. Consider the following distribution :

i. Calculate the frequency of class 30 - 40.
ii. Calculate the class mark of class 10 - 25
Answer:
i.

So, frequency of the class 30 - 40 is 3.
ii. Class mark \( = \frac{\text{upper limit + lower limit}}{2} \)
Class mark of the class \( 10 - 25 = \frac{10 + 25}{2} = \frac{35}{2} = 17.5 \)

Question. Convert the following data to a less than type distribution.

Answer:

Question. The mean of 'n' observations is \( \bar{x} \), if the first term is increased by 1, second by 2 and so on. What will be the new mean?
Answer: Mean of the series \( = \bar{x} \)
According to the question, terms of new series-
I term + 1
II term + 2
III term + 3
n terms + n
The mean of the new numbers is \( \bar{x} + \frac{\frac{n(n+1)}{2}}{n} = \bar{x} + \frac{(n+1)}{2} \)

Question. Find the mean of the following frequency distribution :

Answer:

\( \sum fx = 630, \sum f = 40 \)
Mean \( = \frac{\sum fx}{\sum f} = \frac{630}{40} = 15.75 \)

Question. Following is the distribution of marks of 70 students in a periodical test:

Draw a cumulative frequency curve for the above data.
Answer: We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve.

Question. If the mean of the following distribution is 27, find the value of p.

Answer:

Given mean \( = 27 \)
\( Mean = \frac{sum}{N} \Rightarrow \frac{1245 + 15p}{43 + p} = 27 \)
\( 1245 + 15p = 1161 + 27p \)
\( 12p = 84 \)
\( P = 7 \)

Question. If the mean of the following data is 14.7, find the values of p and q.

Answer:

Given, \( \sum f_i = 40 \),
\( \Rightarrow 26 + p + q = 40 \).
\( \Rightarrow p + q = 40 - 26 \)
\( \Rightarrow p + q = 14 \)...(i)
\( \therefore \text{Mean}, \bar{x} = \frac{\sum x_i f_i}{\sum f_i} \)
\( \Rightarrow 14.7 = \frac{408 + 9p + 27q}{40} \)
\( \Rightarrow 588 = 408 + 9p + 27q \)
\( \Rightarrow 588 - 408 = 9p + 27q \)
\( \Rightarrow 180 = 9p + 27q \)
\( \Rightarrow p + 3q = 20 \)......(ii)
Subtracting eq(i) from eq (ii),
\( 2q = 6 \)
\( \Rightarrow q = 3 \)
Putting this value of q in eq(i),
\( p = 14 - q \)
\( p = 14 - 3 \)
\( p = 11 \)
Hence, \( p = 11, q = 3 \)

Question. Find the mean marks per student, using assumed-mean method:

Answer: the assumed mean is 25.

we know that, \( \text{mean} = A + \frac{\sum (f_i \times d_i)}{\sum f_i} \)
From table, \( \sum f_i = 100 \) and \( \sum (f_i \times d_i) = 300 \)
\( = \left( 25 + \frac{300}{100} \right) \)
\( = 25 + 3 = 28 \)

Question. In a health check-up, the number of heart beats of 40 women are recorded in the following table:

Find the mean of data.
Answer:

Here, class intervals are not continuous . But mid - value \( x_i \) of each class interval would be same either class interval is continuous or not continuous. So, we solve it without making it continuous. Also, \( x_i \)'s are larger, so we apply step-deviation method. Here, class width \( (h) = 5 \).
Table for the given data is:
Now, we have \( a = 72, h = 5, \sum f_i = 40, \sum f_i u_i = 22 \).
\( \therefore \) By step- deviation method,
\( \text{Mean}(\bar{x}) = a + \frac{\sum f_i u_i}{\sum f_i} \times h \)
\( = 72 + \frac{22}{40} \times 5 \)
\( = 72 + \frac{22}{8} \)
\( = 72 + 2.75 \)
\( = 74.75 \)
Mean of given data is 74.75.

Question. Following is the age distribution of a group of students. Draw the cumulative frequency curve of 'less than' type and hence obtain the median value.

Answer:
'less than' type :

And, plot the points A(5, 36), B(6, 78), C(7, 130), D(8, 190), E(9, 258), F(10, 342), G(11, 438), H(12, 520), I(13, 586), J(14, 634), K(15, 684) and L(16, 700). We join freehand these points to get the cumulative frequency curve. On a graph paper, we take the scale.
Scale:
Along x - axis, 5 small divisions = 1
Along the y - axis, 1 small division = 10
Here, \( N = 700 \Rightarrow \frac{N}{2} = 350 \)
Take a point P(0, 350) on the y-axis and draw PQ || x-axis, meeting the curve at Q. Draw QM \( \perp \) x-axis, intersecting the x-axis at M whose coordinates are (10, 0). Hence, median = 10 years.

Question. For the following distribution, calculate mean using all suitable methods:

Answer:
By direct method

\( \text{Mean} = \frac{sum}{N} \)
\( = \frac{848}{64} \)
\( = 13.25 \)
By assuming mean method
Let the assumed mean (A) = 6.5

\( \text{Mean} = A + \frac{sum}{N} = 6.5 + 432/64 \)
\( = 6.5 + 6.75 \)
\( = 13.25 \)

Question. The following table gives the height of trees:

Draw 'less than' ogive and 'more than' ogive.
Answer:
By less than method:

Plot the points (7,26) , (14,57) , (21,92) , (28,134) , (35,216) , (42,287) , (49,341) , (56,360) by taking upper class limit over the x-axis and cumulative frequency over the y-axis.
By more than method:

plot (0,360) , (7,334) , (14,303) , (21,268) , (28,226) , (35,144) , (42,73) , (49,19) by taking lower class limit over the x-axis and cumulative frequency over the y-axis.