CBSE Class 10 Statistics Sure Shot Questions Set 03

CBSE Class 10 Statistics Sure Shot Questions Set C. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Very Short Answer Type Questions (VSA)

Question. If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, then find the median.
Answer: We have, 3 Median = Mode + 2 Mean
= 55.2 + 2(53.4) = 55.2 + 106.8 = 162
\(\Rightarrow\) Median = 162/3 = 54

Question. If the class marks of a distribution are 5, 15, 25, 35, ...., then find second and sixth class intervals.
Answer: If \( h \) is the class size i.e., difference of two consecutive class marks and \( x \) is the class mark of a class-interval, then class interval is \( \left(x - \frac{h}{2}\right) - \left(x + \frac{h}{2}\right) \)
Second class interval is \( \left(15 - \frac{10}{2}\right) - \left(15 + \frac{10}{2}\right) \) i.e., 10–20.
Also, sixth class interval is \( \left(55 - \frac{10}{2}\right) - \left(55 + \frac{10}{2}\right) \) i.e., 50–60.

Question. For a certain frequency distribution, if \( \sum f_i = 50 \) and \( \sum f_i x_i = 2550 \), then what is the mean of the distribution?
Answer: Mean of the distribution = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{2550}{50} = 51 \).

Question. If \( \sum f_i = 11 \), \( \sum f_i x_i = 2p + 52 \) and the mean of the distribution is 6, then find the value of p.
Answer: We have, \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow 6 = \frac{2p + 52}{11} \)
\(\Rightarrow 2p = 66 - 52 \Rightarrow 2p = 14 \Rightarrow p = 7 \)

Question. In the graphical representation of a frequency distribution, if the distance between mode and mean is k times the distance between median and mean, then find the value of k.
Answer: According to the question, Mode – Mean = k(Median – Mean)
\(\Rightarrow\) Mode = k Median – k Mean + Mean
\(\Rightarrow\) Mode = k Median – (k – 1) Mean
Comparing it with empirical relationship \( \text{Mode} = 3 \text{Median} - 2 \text{Mean} \), we have k = 3.

Question. Find the mean of the measures of all the interior angles of a pentagon.
Answer: Mean = \( \frac{\text{Sum of all interior angles of pentagon}}{5} \)
= \( \frac{(3 \times 180^{\circ})}{5} = \frac{540^{\circ}}{5} = 108^{\circ} \)

Question. Find the class marks of classes 20-25 and 45-55.
Answer: The class mark of class 20–25 = \( \frac{20 + 25}{2} = 22.5 \)
The class mark of class 45–55 = \( \frac{45 + 55}{2} = 50 \)

Question. Find the mean of the squares of the first n natural numbers.
Answer: Mean of \( 1^2, 2^2, 3^2, ........., n^2 \)
= \( \frac{1^2 + 2^2 + 3^2 + ......... + n^2}{n} = \frac{n(n + 1)(2n + 1)}{6n} \)
= \( \frac{(n + 1)(2n + 1)}{6} \)

Question. While computing mean of a grouped data, what assumption do we make ?
Answer: While computing mean of a grouped data, we assume that frequencies of various classes are centred at the respective class marks.

Question. Find the median of the following observation: 3, 6, 7, 11, 13, 17, 18, 25
Answer: Ascending order of given observations : 3, 6, 7, 11, 13, 17, 18, 25
Here, n = 8 (Even)
\(\therefore \text{Median} = \frac{(4^{th} + 5^{th})\text{observation}}{2} = \frac{11 + 13}{2} = \frac{24}{2} = 12 \)

Short Answer Type Questions 

Question. Find the mean of the following distribution :

Answer: The frequency distribution table from the given data can be drawn as :

\(\therefore \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{326}{40} = 8.15 \)

Question. Find the mean of the following data:

Answer: The frequency distribution table from the given data can be drawn as :

\(\therefore \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{426}{80} = 5.325 \)

Question. Find the mode of the following data :

Answer: From the given data, we observe that, highest frequency is 20, which lies in the class-interval 40-50.
\(\therefore l = 40, f_1 = 20, f_0 = 12, f_2 = 11, h = 10 \)
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 40 + \left( \frac{20 - 12}{40 - 12 - 11} \right) \times 10 = 40 + \frac{80}{17} = 40 + 4.7 = 44.7 \)

Question. Find the mode of the following distribution :

Answer: From the given data, we observed that, highest frequency is 12, which lies in the class-interval 30-40.
\(\therefore\) 30-40 is the modal class.
\( l = 30, f_1 = 12, f_0 = 7, f_2 = 5, h = 10 \)
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 30 + \left( \frac{12 - 7}{2 \times 12 - 7 - 5} \right) \times 10 = 30 + \left( \frac{5}{24 - 12} \right) \times 10 \)
= \( 30 + \frac{50}{12} = 30 + 4.17 = 34.17 \)

Question. In a class test, 50 students obtained marks are as follows. Find the modal class and the median class.

Answer: The frequency distribution table for the given data can be drawn as :

\(\therefore\) 40-60 is the modal class as it has highest frequency.
Also, \( \frac{N}{2} = \frac{50}{2} = 25 \)
The c.f. just greater than 25 lies in the interval 40-60. Hence the median class is 40-60.

Question. The table below shows the salaries of 280 persons :

Calculate the median salary of the data.
Answer: The cumulative frequency table :

Now, we have \( N = 280 \Rightarrow \frac{N}{2} = \frac{280}{2} = 140 \)
Since, the cumulative frequency just greater than 140 is 182.
\(\therefore\) The median class is 10-15.
and also \( l = 10, c.f. = 49, f = 133 \text{ and } h = 5 \)
\(\therefore \text{Median} = l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \)
= \( 10 + \left[ \frac{140 - 49}{133} \right] \times 5 = 10 + \frac{91 \times 5}{133} = 10 + 3.42 = 13.42 \)
\(\therefore \text{Median salary} = 13.42 \text{ thousand.} \)

Question. Find out the mode for the following data showing frequency with which profits are made:

Answer: From the given data, we observe that the highest frequency is 83, which lies in the class interval 3-4.
\(\therefore\) Modal class is 3-4.
So, \( l = 3, h = 1, f_1 = 83, f_0 = 0, f_2 = 27 \)
\(\therefore \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 3 + \left( \frac{83 - 0}{2 \times 83 - 0 - 27} \right) \times 1 = 3 + \frac{83}{166 - 27} = 3 + \frac{83}{139} \)
= 3 + 0.5971 = 3.5971

Short Answer Type Questions 

Question. The mean weight of 150 students in a class is 60 kg. The mean weight of boys is 70 kg while that of girls is 55 kg. Find the number of boys and girls in the class.
Answer: Total number of students = 150
Mean weight = 60 kg
\(\therefore\) Total weight of 150 students = 150 \(\times\) 60 = 9000 kg
Let the total number of boys be x.
\(\therefore\) Total number of girls = 150 – x
Mean weight of boys = 70 kg
\(\therefore\) Total weight of boys = 70 \(\times\) x = 70x kg
Mean weight of girls = 55 kg
\(\therefore\) Total weight of girls = (150 – x) 55 kg
Now, Total weight = Weight of boys + Weight of girls
\(\Rightarrow 9000 = 70x + (150 – x) 55 \)
\(\Rightarrow 9000 = 70x + 150 \times 55 – 55 x \)
\(\Rightarrow 9000 – 8250 = 70x – 55 x \)
\(\Rightarrow 750 = 15x \Rightarrow x = 50 \)
\(\therefore\) Number of boys = 50 and number of girls = 100

Question. Find the mean of the following distribution:

Answer: Constructing the frequency table :

\(\therefore \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{522}{36} = 14.5 \)

Question. Find the median of the following data :

Answer: The frequency distribution table from the given data can be drawn as :

Here, N = 100, \( \frac{N}{2} = 50 \), which lies in the class interval 20-30.
\(\therefore\) Median class is 20-30.
Median = \( l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \)
= \( 20 + \left[ \frac{50 - 24}{36} \right] \times 10 = 20 + 7.22 = 27.22 \)

Question. Find the modal literacy rate for the following table which gives the literacy rate of 40 cities :

Answer: From the given data, we observe that, highest frequency is 10, which lies in the class interval 50-60.
\(\therefore l = 50, f_1 = 10, f_0 = 7, h = 10, f_2 = 6 \)
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
= \( 50 + \left( \frac{10 - 7}{2 \times 10 - 7 - 6} \right) \times 10 = 50 + \frac{30}{7} = 50 + 4.29 = 54.29 \)

Question. If the mode of the given data is 340, find the missing frequency x for the following data :

Answer: Here, mode is 340, which lies in the interval 300-400. \(\therefore\) Modal class is 300-400.
Now, Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \)
\(\Rightarrow 340 = 300 + \left( \frac{20 - x}{2 \times 20 - x - 14} \right) \times 100 \)
\(\Rightarrow 340 - 300 = \left( \frac{20 - x}{26 - x} \right) \times 100 \)
\(\Rightarrow 6x = 96 \Rightarrow x = 16 \)

Question. Find the missing frequencies \( f_1, f_2 \text{ and } f_3 \) in the following frequency distribution, when it is given that \( f_2 : f_3 = 4 : 3 \), and mean = 50.

Answer: Let \( f_2 = 4x \text{ and } f_3 = 3x \).
The frequency distribution table can be drawn as :

Here, 17 + \( f_1 \) + 4x + 3x + 19 = 120
\(\Rightarrow f_1 + 7x = 84 \Rightarrow f_1 = 84 – 7x \quad \quad ...(i) \)
Also, \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow 50 = \frac{1880 + 30 f_1 + 410x}{120} \)
\(\Rightarrow 6000 = 1880 + 30 f_1 + 410x \Rightarrow 30f_1 + 410x = 4120 \)
\(\Rightarrow 3 f_1 + 41x = 412 \Rightarrow 3(84 – 7x) + 41x = 412 \text{ [Using (i)]} \)
\(\Rightarrow 252 – 21x + 41x = 412 \Rightarrow 20x = 160 \Rightarrow x = 8 \)
When x = 8, \( f_1 = 84 – 7 \times 8 = 84 – 56 = 28 \), \( f_2 = 4 \times 8 = 32 \text{ and } f_3 = 3 \times 8 = 24 \).
\(\therefore\) The missing frequency are \( f_1 = 28, f_2 = 32 \text{ and } f_3 = 24 \).

Question. Find the value of p, if the mean of the following distribution is 7.5.

Answer: Let us construct the following table for the given data.

\(\therefore \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \Rightarrow 7.5 = \frac{303 + 9p}{41 + p} \)
\(\Rightarrow 7.5 \times (41 + p) = 303 + 9p \Rightarrow 307.5 + 7.5p = 303 + 9p \)
\(\Rightarrow 9p – 7.5p = 307.5 – 303 \Rightarrow 1.5 p = 4.5 \Rightarrow p = 3 \)

Long Answer Type Questions 

Question. Weekly income of 600 families is given below. Find the median.

Answer: Cumulative frequency table for the given data is as follows:

Here, n = 600 \(\Rightarrow \frac{n}{2} = 300 \)
Cumulative frequency just greater than 300 is 440 and corresponding interval is 1000 – 2000.
\(\therefore\) Median class is 1000–2000.
So, l = 1000, c.f. = 250, f = 190, h = 1000
\(\therefore \text{Median} = l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h = 1000 + \left( \frac{300 - 250}{190} \right) \times 1000 \)
= 1000 + 263.158 = 1263.158

Question. The median of the distribution given below is 14.4. Find the values of x and y, if the sum of the frequencies is 20.

Answer: Cumulative frequency table for the given data is as follows :

Here, n = 20 \(\Rightarrow \frac{n}{2} = 10 \)
Given, median = 14.4, which lies in the interval 12-18.
\(\therefore\) Median class is 12 - 18. So, l = 12, c.f. = 4 + x, f = 5, h = 6
\(\therefore \text{Median} = l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h \Rightarrow 14.4 = 12 + \left( \frac{10 - (4 + x)}{5} \right) \times 6 \)
\(\Rightarrow 2.4 = \frac{6 - x}{5} \times 6 \Rightarrow 2 = 6 - x \Rightarrow x = 4 \)
Also, given 10 + x + y = 20
\(\Rightarrow 4 + y + 10 = 20 \Rightarrow y = 6 \)

Question. Find the mean, mode and median of the following frequency distribution:

Answer: Constructing the cumulative frequency table:

\(\therefore \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2850}{80} = 35.625 \)
The maximum frequency is 20, which lies in the interval 30-40. \(\therefore\) modal class is 30-40.
Mode = \( l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 30 + \left( \frac{20 - 15}{2 \times 20 - 15 - 12} \right) \times 10 = 30 + \frac{50}{13} = 33.85 \)
Now, n = 80 \(\Rightarrow \frac{n}{2} = 40 \). Median class is 30-40.
Median = \( l + \left( \frac{\frac{n}{2} - c.f.}{f} \right) \times h = 30 + \left( \frac{40 - 30}{20} \right) \times 10 = 35 \).

Question. Find the mean age (in years) from the frequency distribution given below :

Answer: Constructing the table with modified classes:

\(\therefore \text{Mean age} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2755}{70} = 39.35 \text{ years}\).

Question. If the median of the distribution given below is 27. Find the value of x and y.

Answer: Preparing the cumulative frequency table:

Here, N = 68, Median = 27, which lies in the interval 20-30.
\(\therefore \text{Median} = l + \left( \frac{\frac{N}{2} - c.f.}{f} \right) \times h \Rightarrow 27 = 20 + \left( \frac{34 - (5 + x)}{20} \right) \times 10 \)
\(\Rightarrow 7 = \frac{29 - x}{2} \Rightarrow 14 = 29 - x \Rightarrow x = 15 \)
Also, 47 + x + y = 68 \(\Rightarrow 47 + 15 + y = 68 \Rightarrow y = 6 \).

Question. In a village, number of members in 50 families are given in the following frequency distribution :

Find the mode and mean of the above data.
Answer: Mode: Maximum frequency is 10, in class interval 7-9.
Modal class is 7-9. l = 7, h = 2, f1 = 10, f0 = 6, f2 = 5.
Mode = \( 7 + \left[ \frac{10 - 6}{2 \times 10 - 6 - 5} \right] \times 2 = 7 + \frac{8}{9} = 7.889 \).
Mean: Sum fixi = 478, Sum fi = 50.
Mean = \( \frac{478}{50} = 9.56 \).

Question. Find the median of the following frequency distribution:

Answer: Transform table to continuous form:

n = 98 \(\Rightarrow\) n/2 = 49. Median class is 89.5–99.5.
l = 89.5, h = 10, f = 30, c.f. = 40.
Median = \( 89.5 + \left( \frac{49 - 40}{30} \right) \times 10 = 89.5 + 3 = 92.5 \).

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