CBSE Class 10 Mathematics Statistics Assignment Set 05

LEVEL - 1

Question 1. What is the mean of \(1^{\text{st}}\) ten prime numbers?
Answer: The first ten prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Number of observations (\(n\)) = 10
Sum of observations = \(2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129
\text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{129}{10} = 12.9\)
In simple words: Add the first ten prime numbers and divide the total by 10 to get the mean.

Exam Tip: Always list out the prime numbers carefully starting from 2 (1 is neither prime nor composite) to avoid missing any term.

Question 2. What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect?
Answer: Median
In simple words: The x-coordinate (abscissa) of the intersection point of less than and more than cumulative frequency curves gives the median.

Exam Tip: Remember that 'abscissa' means the x-coordinate, which corresponds to the value of the median on the horizontal axis.

Question 3. If the mode of a data is 45 and mean is 27, then median is ___________.
Answer: Using the empirical relationship between mean, median, and mode:
\(3 \text{ Median} = \text{Mode} + 2 \text{ Mean}\)
\(3 \text{ Median} = 45 + 2(27)\)
\(3 \text{ Median} = 45 + 54 = 99
\text{Median} = \frac{99}{3} = 33\)
In simple words: Apply the formula connecting mean, median, and mode to calculate the missing value.

Question 4. Find the mode of the following:
\(X_i\): 35, 38, 40, 42, 44
\(f_i\): 5, 9, 10, 7, 2
Answer: The observation with the highest frequency is 40, which has a corresponding frequency of 10.
Therefore, the mode = 40.
In simple words: Mode is the value that occurs most often. Look for the highest frequency in the given table.

Exam Tip: For discrete frequency distributions, simply identify the variate having the maximum frequency.

Question 5. Write the median class of the following distribution.
Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Frequency: 4, 4, 8, 10, 12, 8, 4
Answer:
Let us construct the cumulative frequency table:

Total frequency \(n = 50\)
\(\frac{n}{2} = \frac{50}{2} = 25\)
The cumulative frequency just greater than 25 is 26, which corresponds to the class interval 30-40.
Therefore, the median class is 30-40.
In simple words: Find half of the total frequency (\(n/2\)) and locate the class whose cumulative frequency is just greater than or equal to this value.

Exam Tip: Always double check your cumulative frequency addition before finding \(\frac{n}{2}\).

LEVEL - 2

Question 1. Calculate the mean of the following distribution:
Class interval: 50-60, 60-70, 70-80, 80-90, 90-100
Frequency: 8, 6, 12, 11, 13
Answer:
Let us use the assumed mean method. Let assumed mean \(a = 75\) and class size \(h = 10\).

\(\text{Mean } (\overline{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h\)
\(\text{Mean} = 75 + \left(\frac{15}{50}\right) \times 10\)
\(\text{Mean} = 75 + 3 = 78\)
In simple words: Use the step-deviation method to simplify large numbers and calculate the mean easily.

Exam Tip: Choosing the middle class mark as the assumed mean \(a\) makes calculations much simpler by generating smaller integers.

Question 2. Find the mode of the following frequency distribution:
Marks: 10-20, 20-30, 30-40, 40-50, 50-60
No. of students: 12, 35, 45, 25, 13
Answer:
The highest frequency is 45, which corresponds to the class interval 30-40.
Therefore, the modal class is 30-40.
Lower limit of modal class (\(l\)) = 30
Size of class interval (\(h\)) = 10
Frequency of modal class (\(f_1\)) = 45
Frequency of preceding class (\(f_0\)) = 35
Frequency of succeeding class (\(f_2\)) = 25
\(\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h\)
\(\text{Mode} = 30 + \left(\frac{45 - 35}{2(45) - 35 - 25}\right) \times 10\)
\(\text{Mode} = 30 + \left(\frac{10}{90 - 60}\right) \times 10\)
\(\text{Mode} = 30 + \left(\frac{10}{30}\right) \times 10\)
\(\text{Mode} = 30 + \frac{10}{3} = 30 + 3.33 = 33.33\)
In simple words: Identify the modal class with the highest frequency and apply the standard mode formula.

Exam Tip: Always double check that your modal class has the maximum frequency before substituting values into the formula.

Question 3. Find the median of the following distribution:
Class interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
Frequency: 5, 8, 20, 15, 7, 5
Answer:
Let us construct the cumulative frequency table:

Total frequency \(n = 60\)
\(\frac{n}{2} = \frac{60}{2} = 30\)
The cumulative frequency just greater than 30 is 33, corresponding to the class interval 20-30.
Median class = 20-30
Lower limit (\(l\)) = 20
Cumulative frequency of preceding class (\(cf\)) = 13
Frequency of median class (\(f\)) = 20
Class size (\(h\)) = 10
\(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
\(\text{Median} = 20 + \left(\frac{30 - 13}{20}\right) \times 10\)
\(\text{Median} = 20 + \left(\frac{17}{20}\right) \times 10\)
\(\text{Median} = 20 + \frac{17}{2} = 20 + 8.5 = 28.5\)
In simple words: Find the median class using \(n/2\) and substitute the values into the median formula.

Exam Tip: Take care to use the cumulative frequency of the class *preceding* the median class, not the median class itself.

Question 4. A class teacher has the following absentee record of 40 students of a class for the whole term.
No. of days: 0-6, 6-10, 10-14, 14-20, 20-28, 28-38, 38-40
No. of students: 11, 10, 7, 4, 4, 3, 1
Write the above distribution as less than type cumulative frequency distribution.
Answer:
To write a less than type cumulative frequency distribution, we use the upper class limits and add the frequencies successively.

In simple words: Cumulative frequencies are obtained by continuously adding up the frequencies for each upper class limit.

Exam Tip: The final cumulative frequency must always equal the total number of observations given in the problem.

LEVEL - 3

Question 1. If the mean distribution is 25:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 18, 15, P, 6
Then find p.
Answer:
Let us set up the frequency distribution table with class marks (\(x_i\)) and \(f_i x_i\):

Given mean (\(\overline{x}\)) = 25
\(\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}\)
\(25 = \frac{940 + 35P}{44 + P}\)
\(25(44 + P) = 940 + 35P\)
\(1100 + 25P = 940 + 35P\)
\(1100 - 940 = 35P - 25P\)
\(160 = 10P\)
\(P = 16\)
In simple words: Use the direct mean formula and solve the linear equation for the unknown variable \(P\).

Exam Tip: Be careful with algebraic multiplication when expanding terms like \(25(44 + P)\).

Question 2. Find the mean of the following frequency distribution using step deviation method:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 7, 12, 13, 10, 8
Answer:
Let assumed mean \(a = 25\) and class size \(h = 10\).

\(\text{Mean } (\overline{x}) = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h\)
\(\text{Mean} = 25 + \left(\frac{0}{50}\right) \times 10 = 25 + 0 = 25\)
In simple words: When the sum of step-deviations \(\sum f_i u_i\) turns out to be zero, the mean is exactly equal to the assumed mean.

Exam Tip: A zero value for \(\sum f_i u_i\) indicates that your assumed mean is precisely the true mean of the distribution.

Question 3. Find the value of p if the median of the following frequency distribution is 50:
Class: 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90
Frequency: 25, 15, P, 6, 24, 12, 8
Answer:
Let us construct the cumulative frequency table:

Given median = 50, which falls in the class interval 50-60.
Therefore, median class = 50-60.
Lower limit (\(l\)) = 50
Frequency of median class (\(f\)) = 6
Cumulative frequency of preceding class (\(cf\)) = \(40 + P\)
Class size (\(h\)) = 10
\(\frac{n}{2} = \frac{90 + P}{2}\)
\(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
\(50 = 50 + \left(\frac{\frac{90 + P}{2} - (40 + P)}{6}\right) \times 10\)
\(0 = \frac{\frac{90 + P - 80 - 2P}{2}}{6} \times 10\)
\(0 = \frac{10 - P}{12} \times 10\)
\(10 - P = 0 \implies P = 10\)
In simple words: Locate the median class using the given median value and apply the median formula to solve for \(P\).

Exam Tip: When the median is explicitly given, immediately identify its corresponding class interval to determine \(l\), \(f\), and \(cf\).

Question 4. Find the median of the following data:
Marks: Less Than 10, Less Than 30, Less Than 50, Less Than 70, Less Than 90, Less Than 110, Less Than 130, Less than 150
Frequency: 0, 10, 25, 43, 65, 87, 96, 100
Answer:
Since the given frequencies are already cumulative frequencies, let us convert them into standard class intervals and frequencies:

Total frequency \(n = 100\)
\(\frac{n}{2} = \frac{100}{2} = 50\)
The cumulative frequency just greater than 50 is 65, which corresponds to the class interval 60-80.
Median class = 60-80
Lower limit (\(l\)) = 60
Class size (\(h\)) = 20
Frequency (\(f\)) = 22
Cumulative frequency of preceding class (\(cf\)) = 43
\(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
\(\text{Median} = 60 + \left(\frac{50 - 43}{22}\right) \times 20\)
\(\text{Median} = 60 + \left(\frac{7}{22}\right) \times 20\)
\(\text{Median} = 60 + \frac{140}{22} = 60 + 6.36 = 76.36\)
In simple words: Convert "less than" cumulative frequencies into regular class frequencies before applying the median formula.

Exam Tip: "Less than" values represent upper class limits. Subtract consecutive cumulative frequencies to obtain individual class frequencies.

LEVEL - 4

Question 1. The mean of the following frequency distribution is 57.6 and the sum of the observations is 50. Find the missing frequencies \(f_1\) and \(f_2\).
Class: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120, Total
Frequency: 7, \(f_1\), 12, \(f_2\), 8, 5, 50
Answer:
Let us set up the frequency distribution table with class marks and products:

From the total frequency:
\(32 + f_1 + f_2 = 50 \implies f_1 + f_2 = 18 \quad \text{--- (Equation 1)}\)
Given mean = 57.6
\(\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}\)
\(57.6 = \frac{1940 + 30f_1 + 70f_2}{50}\)
\(57.6 \times 50 = 1940 + 30f_1 + 70f_2\)
\(2880 = 1940 + 30f_1 + 70f_2\)
\(30f_1 + 70f_2 = 2880 - 1940\)
\(30f_1 + 70f_2 = 940 \implies 3f_1 + 7f_2 = 94 \quad \text{--- (Equation 2)}\)
From Equation 1, \(f_1 = 18 - f_2\). Substituting into Equation 2:
\(3(18 - f_2) + 7f_2 = 94\)
\(54 - 3f_2 + 7f_2 = 94\)
\(4f_2 = 94 - 54 = 40 \implies f_2 = 10\)
Substitute \(f_2 = 10\) into Equation 1:
\(f_1 + 10 = 18 \implies f_1 = 8\)
Therefore, \(f_1 = 8\) and \(f_2 = 10\).
In simple words: Form two equations using the total frequency sum and the mean formula, then solve them simultaneously.

Exam Tip: Problems with two missing frequencies always yield two simultaneous linear equations in two variables.

Question 2. The following distribution give the daily income of 65 workers of a factory:
Daily income (in Rs): 100-120, 120-140, 140-160, 160-180, 180-200
No. of workers: 14, 16, 10, 16, 9
Convert the above to a more than type cumulative frequency distribution and draw its ogive.
Answer:
Let us construct the more than type cumulative frequency table using lower limits:

In simple words: For "more than" ogives, start with the total frequency and subtract each class frequency successively from top to bottom.

Exam Tip: Plot lower class limits on the x-axis and the corresponding cumulative frequencies on the y-axis to draw a "more than" ogive curve.

Question 3. Draw a less than type and more than type ogives for the following distribution on the same graph. Also find the median from the graph.
Marks: 30-39, 40-49, 50-59, 60-69, 70-79, 80-89, 90-99
No. of students: 14, 6, 10, 20, 30, 8, 12
Answer:
First, convert the discontinuous class intervals into continuous class intervals by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits:

To find the median graphically, plot both less than and more than ogives on the same coordinate axes. The x-coordinate of their intersection point represents the median.
In simple words: Make class intervals continuous first, prepare both cumulative frequency tables, and locate the median at the intersection of both ogives.

Exam Tip: Always check if class intervals are inclusive or exclusive. Convert inclusive classes (like 30-39) to exclusive classes (29.5-39.5) before drawing ogives.

SELF - EVALUATION

Question 1. What is the value of the median of the data using the graph in figure of less than ogive and more than ogive?
Answer: Looking at the given intersection point of the less than and more than ogive curves, the abscissa (x-coordinate) gives the median value.
In simple words: Drop a perpendicular from the intersection point of the two curves to the horizontal axis to read off the median.

Exam Tip: The y-coordinate at the intersection point is always equal to \(\frac{n}{2}\).

Question 2. If mean = 60 and median = 50, then find mode using empirical relationship.
Answer: Using the empirical formula:
\(\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}\)
\(\text{Mode} = 3(50) - 2(60)\)
\(\text{Mode} = 150 - 120 = 30\)
In simple words: Substitute the given values of median and mean into the empirical formula to calculate the mode.

Exam Tip: Remember the two equivalent forms: \(3 \text{ Median} = \text{Mode} + 2 \text{ Mean}\) and \(\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}\).

Question 3. Find the value of p, if the mean of the following distribution is 18.
Variate (\(x_i\)): 13, 15, 17, 19, 20+p, 23
Frequency (\(f_i\)): 8, 2, 3, 4, 5p, 6
Answer:
Let us construct the table for calculating \(\sum f_i x_i\):

Given mean = 18
\(\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}\)
\(18 = \frac{399 + 100p + 5p^2}{23 + 5p}\)
\(18(23 + 5p) = 399 + 100p + 5p^2\)
\(414 + 90p = 399 + 100p + 5p^2\)
\(5p^2 + 10p - 15 = 0\)
Divide by 5: \(p^2 + 2p - 3 = 0\)
\((p + 3)(p - 1) = 0\)
Since frequency cannot be negative, \(p = 1\).
In simple words: Multiply each variate by its frequency, set up the mean equation, and solve the resulting quadratic equation.

Exam Tip: Always discard negative values for frequency or variables representing counts.

Question 4. Find the mean, mode and median for the following data:
Classes: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Frequency: 5, 8, 15, 20, 14, 8, 5
Answer:
Let us construct the complete statistical table:

1. Mean:
\(\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2615}{75} = 34.87\)
2. Mode:
Highest frequency is 20, corresponding to class 30-40.
\(l = 30, h = 10, f_1 = 20, f_0 = 15, f_2 = 14\)
\(\text{Mode} = 30 + \left(\frac{20 - 15}{2(20) - 15 - 14}\right) \times 10 = 30 + \left(\frac{5}{11}\right) \times 10 = 30 + 4.55 = 34.55\)
3. Median:
\(\frac{n}{2} = \frac{75}{2} = 37.5\), which gives median class 30-40.
\(l = 30, cf = 28, f = 20, h = 10\)
\(\text{Median} = 30 + \left(\frac{37.5 - 28}{20}\right) \times 10 = 30 + \left(\frac{9.5}{2}\right) = 30 + 4.75 = 34.75\)
In simple words: Create a single table with class marks and cumulative frequencies to calculate mean, mode, and median step by step.

Exam Tip: Organize your table neatly with separate columns for \(x_i\), \(f_i x_i\), and \(cf\) to prevent calculation errors.

Question 5. The median of the following data is 52.5. Find the value of x and y, if the total frequency is 100.
Class Interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100
Frequency: 2, 5, x, 12, 17, 20, y, 9, 7, 4
Answer:
Let us construct the cumulative frequency table:

From total frequency:
\(76 + x + y = 100 \implies x + y = 24 \quad \text{--- (Equation 1)}\)
Given median = 52.5, which lies in the class interval 50-60.
Median class = 50-60
\(l = 50, f = 20, cf = 36 + x, h = 10, \frac{n}{2} = 50\)
\(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
\(52.5 = 50 + \left(\frac{50 - (36 + x)}{20}\right) \times 10\)
\(2.5 = \frac{14 - x}{2}\)
\(5 = 14 - x \implies x = 9\)
Substitute \(x = 9\) into Equation 1:
\(9 + y = 24 \implies y = 15\)
Therefore, \(x = 9\) and \(y = 15\).
In simple words: Use the total frequency sum and the median formula with the given median value to find the two unknowns.

Exam Tip: The median class is determined directly by checking which interval contains the given median value (52.5).

Question 6. Draw 'less than ogive' and 'more than ogive' for the following distribution and hence find its median.
Classes: 20-30, 30-40, 40-50, 50-60, 60-70, 70-80, 80-90
Frequency: 10, 8, 12, 24, 6, 25, 15
Answer:
Let us prepare the cumulative frequency table for both less than and more than ogives:

Plot points \((upper\ limit, less\ than\ cf)\) for the less than ogive and \((lower\ limit, more\ than\ cf)\) for the more than ogive. The x-coordinate of their intersection point gives the median.
In simple words: Construct both cumulative frequency columns and plot them on graph paper to find the median at their crossing point.

Exam Tip: Ensure your scale is consistent and clearly marked on both axes when drawing ogives.

Question 7. Find the mean marks for the following data:
Marks: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60, Below 70, Below 80, Below 90, Below 100
No. of students: 5, 9, 17, 29, 45, 60, 70, 78, 83, 85
Answer:
Let us convert the "below" cumulative frequencies into standard class intervals and individual frequencies:

\(\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4135}{85} = 48.65\)
In simple words: Subtract consecutive cumulative frequencies to find individual class frequencies, then apply the direct mean formula.

Exam Tip: "Below" data gives cumulative frequencies directly. Always find differences to get standard frequencies before computing the mean.

Question 8. The following table shows age distribution of persons in a particular region. Calculate the median age.
Age in years: Below 10, Below 20, Below 30, Below 40, Below 50, Below 60, Below 70, Below 80
No. of persons: 200, 500, 900, 1200, 1400, 1500, 1550, 1560
Answer:
Let us convert the given "below" cumulative frequencies into standard class intervals and frequencies:

Total frequency \(n = 1560\)
\(\frac{n}{2} = \frac{1560}{2} = 780\)
The cumulative frequency just greater than 780 is 900, which corresponds to the class interval 20-30.
Median class = 20-30
\(l = 20, f = 400, cf = 500, h = 10\)
\(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
\(\text{Median} = 20 + \left(\frac{780 - 500}{400}\right) \times 10\)
\(\text{Median} = 20 + \left(\frac{280}{400}\right) \times 10\)
\(\text{Median} = 20 + \frac{28}{4} = 20 + 7 = 27\)
In simple words: Find individual frequencies from the given cumulative frequency table and apply the median formula.

Exam Tip: Be careful to use the class interval corresponding to the correct cumulative frequency row.

Question 9. If the median of the following data is 32.5. Find the value of x and y.
Class Interval: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, Total
Frequency: x, 5, 9, 12, y, 3, 2, 40
Answer:
Let us construct the cumulative frequency table:

From total frequency:
\(31 + x + y = 40 \implies x + y = 9 \quad \text{--- (Equation 1)}\)
Given median = 32.5, which lies in the class interval 30-40.
Median class = 30-40
\(l = 30, f = 12, cf = 14 + x, h = 10, \frac{n}{2} = 20\)
\(\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h\)
\(32.5 = 30 + \left(\frac{20 - (14 + x)}{12}\right) \times 10\)
\(2.5 = \frac{6 - x}{12} \times 10\)
\(2.5 \times 12 = 10(6 - x)\)
\(30 = 60 - 10x\)
\(10x = 30 \implies x = 3\)
Substitute \(x = 3\) into Equation 1:
\(3 + y = 9 \implies y = 6\)
Therefore, \(x = 3\) and \(y = 6\).
In simple words: Set up equations using total frequency and the median formula to determine the missing values \(x\) and \(y\).

Exam Tip: Always verify that your solved values of \(x\) and \(y\) satisfy both the total frequency equation and the median interval condition.