CBSE Class 10 Statistics Sure Shot Questions Set 11

Question. Find the class marks of class 10 – 25 and 35 – 55 :
(a) 1.75 and 45
(b) 17.5 and 4.5
(c) 1.75 and 4.5
(d) 17.5 and 45
Answer: (d) 17.5 and 45
Explanation :
We derive the class mark \( x_i \) by the following formula :
\( x_i = \frac{1}{2}(\text{lower limit} + \text{upper limit}) \)
Thus, \( x_i = \frac{1}{2}(10 + 25) \)
\( = \frac{1}{2}(35) = 17.5 \)
and \( x_{ii} = \frac{1}{2}(\text{lower limit} + \text{upper limit}) \)
Thus, \( x_{ii} = \frac{1}{2}(35 + 55) \)
\( = \frac{1}{2}(90) = 45 \)

Question. Write down the median class of the following frequency distribution :
Class Interval: 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50, 50 – 60, 60 – 70
Frequency: 4, 4, 8, 10, 12, 8, 4
(a) 20 – 30
(b) 30 – 40
(c) 40 – 50
(d) 50 – 60
Answer: (b) 30 – 40
Explanation :
Class Interval | Frequency (\( f_i \)) | Cumulative Frequency
0 – 10 | 4 | 4
10 – 20 | 4 | 8
20 – 30 | 8 | 16
30 – 40 | 10 | 26
40 – 50 | 12 | 38
50 – 60 | 8 | 46
60 – 70 | 4 | 50
\( N = \sum f_i = 50 \)
Thus, \( \frac{N}{2} = 25 \)
The cumulative frequency just above 25 is 26.
Hence, the median class is 30 – 40.

Question. Calculate the value of p from the following data :
Class: 0 – 20, 20 – 40, 40 – 60, 60 – 80, 80 – 100
Frequency: 8, 15, p, 12, 5
\( N = \sum f_i = 60 \)
(a) 20
(b) 30
(c) 45
(d) 50
Answer: (a) 20
Explanation :
Given, \( N = \sum f_i = 60 \)
\( \Rightarrow 8 + 15 + p + 12 + 5 = 60 \)
\( \Rightarrow 40 + p = 60 \)
\( \Rightarrow p = 20 \)

Question. In an inclusive series :
(a) The lower class boundary is same as the upper class boundary of the previous class.
(b) The upper class boundary is same as the lower class boundary of the next class.
(c) Both the lower and upper class boundaries are the same.
(d) The lower and upper class boundaries are contained within the class and do not intersect with either the upper boundary of the previous class or the lower boundary of the next class.
Answer: (d) The lower and upper class boundaries are contained within the class and do not intersect with either the upper boundary of the previous class or the lower boundary of the next class.
Explanation :
Inclusive series can be converted into the exclusive series.

Question. \( \sum f_i = 15, \sum f_i x_i = 3p + 36 \) and mean of the distribution is 3, then p will be :
(a) 2
(b) 3
(c) 1
(d) 6
Answer: (b) 3
Explanation :
Mean = \( \frac{\sum f_i x_i}{\sum f_i} \)
\( 3 = \frac{3p + 36}{15} \)
\( 45 = 3p + 36 \)
\( 3p = 9 \)
\( p = 3. \)

Question. If the value of mean and mode are 30 and 15, respectively, then median will be :
(a) 25
(b) 24
(c) 23.5
(d) 26
Answer: (a) 25
Explanation :
We know that
Mode = 3 median – 2 mean
\( \Rightarrow 15 = 3 \text{ median} - 60 \)
\( \Rightarrow 3 \text{ median} = 75 \)
\( \Rightarrow \text{Median} = 25. \)

Question. The mean of the first 10 natural numbers is :
(a) 0
(b) 5.5
(c) 7
(d) 5
Answer: (b) 5.5
Explanation :
\( \because \) The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
\( \therefore \text{Mean} = \frac{1+2+3+4+5+6+7+8+9+10}{10} \)
\( = \frac{55}{10} = 5.5. \)

Question. The relation between mean, median and mode is :
(a) mode = 3 mean – 2 median
(b) mode = 3 median – 2 mean
(c) median = 3 mean – 2 mode
(d) mean = 3 median – 2 mode
Answer: (b) mode = 3 median – 2 mean
Explanation :
This is called an empirical relation between mean, median and mode.

Question. Find the mode of the following data : 0, 5, 5, 1, 6, 4, 3, 0, 2, 5, 5, 6.
(a) 6
(b) 4
(c) 3
(d) 5
Answer: (d) 5
Explanation :
The value 5 has maximum frequency.
Therefore mode = 5.

Question. If median of the following data arranged in an ascending order is 25, then the value of x is : 5, 7, 10, 12, 2x – 8, 2x + 10, 35, 41, 42, 50
(a) 10
(b) 13
(c) 12
(d) 11
Answer: (c) 12
Explanation :
Number of observations (n) = 10
\( \therefore \text{Median} = \frac{1}{2} [ (\frac{n}{2})^{\text{th}} \text{ term} + (\frac{n}{2} + 1)^{\text{th}} \text{ term} ] \)
\( \Rightarrow 25 = \frac{1}{2} [ (2x - 8) + (2x + 10) ] \)
\( \Rightarrow 25 = \frac{1}{2} [ 4x + 2 ] \)
\( \Rightarrow 25 = 2x + 1 \)
\( \Rightarrow x = 12. \)

Question. Find the value of y from the following observations if these are already arranged in ascending order. The median of the given observation is 63. 20, 24, 42, y, y + 2, 73, 75, 80, 99
(a) 61
(b) 79
(c) 45
(d) 65
Answer: (a) 61
Explanation :
Number of observation = 9
Median = \( (\frac{9+1}{2})^{\text{th}} \text{ term} \)
\( = (\frac{10}{2})^{\text{th}} \text{ term} \)
\( = 5^{\text{th}} \text{ term} \)
\( 63 = y + 2 \)
\( y = 61. \)

Question. A student scored the following marks in 6 subjects : 30, 19, 25, 30, 27, 30. Find his modal score :
(a) 20
(b) 25
(c) 30
(d) 26
Answer: (c) 30
Explanation :
As 30 occurs maximum numbers of times.

Question. The mean of the frequency distribution are 28 and 16 respectively. Find the median :
(a) 22.5
(b) 24
(c) 24.5
(d) 26
Answer: (c) 24.5

Question. The median of the following frequency distribution will be :
x: 6, 7, 5, 2, 10, 9, 3
y: 9, 12, 8, 13, 11, 14, 7
(a) 7
(b) 4
(c) 5
(d) 6
Answer: (d) 6
Explanation :
x | f | c.f. | x | f | c.f.
2 | 13 | 13 | 7 | 12 | 49
3 | 7 | 20 | 9 | 14 | 63
5 | 8 | 28 | 10 | 11 | 74
6 | 9 | 37
Here \( \frac{N}{2} = \frac{74}{2} = 37^{\text{th}} \text{ observation} \)
Median = \( 37^{\text{th}} \text{ observation} = 6 \)

Question. If \( \sum f_i = 17, \sum f_i x_i = 4p + 63 \) and mean = 7, then p is :
(a) 14
(b) 13
(c) 12
(d) 11
Answer: (a) 14
Explanation :
Mean = \( \frac{\sum f_i x_i}{\sum f_i} \)
\( \Rightarrow 7 = \frac{4p + 63}{17} \)
\( \Rightarrow p = 14 \)

Question. The wickets taken by a bowler in 10 matches are : 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. Find the mode :
(a) 1
(b) 2
(c) 4
(d) 3
Answer: (b) 2
Explanation :
2 has height frequency of 3. so mode = 2.

Question. What is the mean of the following data :
Class interval: 50–60, 60–70, 70–80, 80–90, 90– 100
F: 8, 6, 12, 11, 13
(a) 78
(b) 68
(c) 48
(d) 58
Answer: (a) 78
Explanation :
Class Interval | f | x | f × x
50 – 60 | 8 | 55 | 440
60 – 70 | 6 | 65 | 390
70 – 80 | 12 | 75 | 900
80 – 90 | 11 | 85 | 935
90 – 100 | 13 | 95 | 1235
Total | \( \sum f = 60 \) | | \( 3900 = \sum f_i x_i \)
Mean = \( \frac{3900}{50} = 78 \).

Question. The mode of a frequency distribution can be determined graphically from ........... .
(a) Bar graph
(b) ogive
(c) Histogram
(d) Pie chart
Answer: (c) Histogram
Explanation :
It is a rectangular blocks used to determine mode graphically.

Question. If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 17, then the value of x is ........... .
(a) 18
(b) 10
(c) 27
(d) 17
Answer: (d) 17
Explanation :
Mode is the value of the data which occurs maximum number of times.

Question. If the mean of observations \( x_1, x_2, x_3, ...... x_n \) is, \( \bar{x} \) then the mean of \( ax_1, ax_2, ax_3, ......, ax_n \) is :
(a) \( \bar{x} \)
(b) \( a + \bar{x} \)
(c) \( a\bar{x} \)
(d) None of the options
Answer: (c) \( a\bar{x} \)

Question. Which of the following cannot be determined graphically?
(a) Mean
(b) Median
(c) Mode
(d) None of the options
Answer: (a) Mean
Explanation :
Mean cannot be determined graphically.

Question. If the mean of first n natural number is 15, then n =
(a) 15
(b) 30
(c) 14
(d) 29
Answer: (d) 29
Explanation :
Mean of first n natural number = 15
\( \Rightarrow \frac{n(n+1)}{2n} = 15 \)
\( \Rightarrow \frac{n+1}{2} = 15 \)
\( \Rightarrow n + 1 = 30 \)
\( \Rightarrow n = 29 \)

Question. For the following distribution :
Class: 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25
Frequency: 10, 15, 12, 20, 9
The sum of lower limits of the median class and modal class is :
(a) 15
(b) 25
(c) 30
(d) 35
Answer: (b) 25
Explanation :
Class | Frequency | Cumulative Frequency
0 – 5 | 10 | 10
5 – 10 | 15 | 25
10 – 15 | 12 | 37
15 – 20 | 20 | 57
20 – 25 | 9 | 66
Now, \( \frac{N}{2} = \frac{66}{2} = 33 \), which lies in the interval 10 – 15. Therefore, lower limit of the median class is 10.
The highest frequency is 20, which ties in the interval 15 – 20. Therefore, lower limit of modal class is 15.
Hence, requried sum is 10 + 15 = 25.

Question. Consider the following frequency distribution :
Class: 0 – 5, 6 – 11, 12 – 17, 18 – 23, 24– 29
Frequency: 13, 10, 15, 8, 11
The upper limit of the median class is :
(a) 11.5
(b) 17.5
(c) 23.5
(d) 29.5
Answer: (b) 17.5
Explanation :
Class | Frequency | Cumulative Frequency
0.5 – 5.5 | 13 | 13
5.5 – 11.5 | 10 | 23
11.5 – 17.5 | 15 | 38
17.5 – 23.5 | 8 | 46
23.5 – 29.5 | 11 | 57
Here, N = 57
Therefore, \( \frac{N}{2} = \frac{57}{2} = 28.5 \)
Median class is 11.5 – 17.5.

Question. If the mean of observation \( x_1, x_2, .... x_n \) is \( \bar{x} \) then the mean of \( x_1 + a, x_2 + a, ..... x_n + a \) is :
(a) \( a\bar{x} \)
(b) \( \bar{x} - a \)
(c) \( \bar{x} + a \)
(d) \( ax \)
Answer: (c) \( \bar{x} + a \)
Explanation :
Mean of observations \( x_1, x_2, ......, x_n \) is \( \bar{x} \)
\( x_1 + x_2 + x_3 + ..... + x_n = n\bar{x} \)
\( x_1 + a + x_2 + a + x_3 + a + ..... x_n + a \)
\( = x_1 + x_2 + x_3 + .... x_n + na \)
Mean of \( (x_1 + x_2 + x_3 + .... + x_n) + na = \frac{n\bar{x} + na}{n} = \bar{x} + a \)

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) \( \bar{x} + n \)
(b) \( \bar{x} + n^2 \)
(c) \( \bar{x} + \frac{n+1}{2} \)
(d) None of the options
Answer: (c) \( \bar{x} + \frac{n+1}{2} \)
Explanation :
Let \( x_1, x_2, x_3, ......., x_n \) be the n observations
Mean = \( \frac{x_1 + x_2 + .... + x_n}{n} = \bar{x} \)
If the first item is increased by 1, second by 2 and so on.
Then, the new observations are \( x_1 + 1, x_2 + 2, x_3 + 3, ..... x_n + n \)
New mean = \( \frac{(x_1+1)+(x_2+2)+(x_3+3)+....+(x_n+n)}{n} \)
\( = \frac{x_1 + x_2 + x_3 + .... + x_n + (1+2+3+....+n)}{n} \)
\( = \frac{n\bar{x}}{n} + \frac{n(n+1)/2}{n} = \bar{x} + \frac{n+1}{2} \)

Question. The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is :
(a) 25
(b) 18
(c) 20
(d) 22
Answer: (c) 20
Explanation :
Arithmetic mean = 24
Mode = 12
But mode = 3 median – 2 mean
\( \Rightarrow 12 = 3 \text{ median} - 2 \times 24 \)
\( \Rightarrow 12 = 3 \text{ median} - 48 \)
\( \Rightarrow 12 + 48 = 3 \text{ median} \)
\( \Rightarrow 3 \text{ median} = 60 \)
Median = 20

Question. While computing mean of grouped data, we assume that the frequencies are :
(a) Evenly distributed over all the classes
(b) Centred at the classmarks of the classes
(c) Centred at the upper limits of the classes
(d) Centred at the lower limits of the classes
Answer: (c) Centred at the upper limits of the classes
Explanation :
In computting the mean of grouped data, the frequencies are centred at the class marks of the classes.

Question. If \( x_i \)'s are the mid-points of the class intervals of grouped data \( f_i \)'s are the corresponding frequencies and \( \bar{x} \) is the mean, then \( S(x_i f_i - \bar{x}) \) equal to :
(a) 0
(b) – 1
(c) 1
(d) 2
Answer: (a) 0
Explanation :
We know that \( \bar{x} = \frac{\sum f_i x_i}{n} \)
\( \Rightarrow n\bar{x} = \sum f_i x_i \)...(i)
\( \Rightarrow \sum(f_i x_i - \bar{x}) = \sum f_i x_i - n\bar{x} \)
\( = n\bar{x} - n\bar{x} = 0 \)[from (i)]

Question. If 35 is removed from the data : 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by :
(a) 2
(b) 1.5
(c) 1
(d) 0.5
Answer: (d) 0.5
Explanation :
Given data = 30, 34, 35, 36, 37, 38, 39, 40
Here n = 8 which is even
Median = \( \frac{36+37}{2} = 36.5 \)
After removing 35, then n = 7
New median = \( 4^{\text{th}} \text{ term} = 37 \)
Increase in median = 37 – 36.5 = 0.5

Question. In the formula \( \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \) finding the mean of grouped data \( d_i \)'s are deviations from :
(a) lower limits of classes
(b) upper limits of classes
(c) mid-points of classes
(d) frequency of the class marks
Answer: (c) mid-points of classes.
Explanation :
We know that, \( d_i = x_i - a \)
i.e., \( d_i \)'s are the deviation from the mid-points of the classes.

Question. The mean of 1, 3, 4, 5, 7, 4, is m. The number 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then p + q =
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (d) 7
Explanation :
Mean of 1, 3, 4, 5, 7, 4 is m
\( \frac{1+3+4+5+7+4}{6} = m \)
\( \Rightarrow \frac{24}{6} = m \)
\( \Rightarrow m = 4 \)
Mean of 3, 2, 2, 4, 3, 3, p is m – 1
\( \Rightarrow \frac{3+2+2+4+3+3+p}{7} = m - 1 \)
\( \Rightarrow 17 + p = 7m - 7 \)
\( \Rightarrow 17 + p = 28 - 7 \)
\( \Rightarrow 17 + p = 21 \)
\( \Rightarrow p = 4 \)
Median of 3, 2, 2, 4, 3, 3, p is q
3, 2, 2, 4, 3, 3, 4 is q
Arranging in order, we get 4, 4, 3, 3, 3, 2, 2
Here n = 7
Median = \( (\frac{7+1}{2})^{\text{th}} \text{ term} = 4^{\text{th}} \text{ term} \)
Median = 3
or q = 3
\( \therefore p + q = 4 + 3 = 7 \)

Question. If the mean of frequency distribution is 8.1 and \( \sum f_i x_i = 132 + 5k, \sum f_i = 20 \), then k =
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (d) 6
Explanation :
Given :
\( \sum f_i x_i = 132 + 5k \),
\( \sum f_i = 20 \)
Mean = 8.1
Then,
Mean = \( \frac{\sum f_i x_i}{\sum f_i} = \frac{132 + 5k}{20} \)
\( \Rightarrow 8.1 = \frac{132 + 5k}{20} \)
\( \Rightarrow 162 = 132 + 5k \)
\( \Rightarrow 5k = 30 \)
\( \Rightarrow k = 6 \)

Question. The mean of 20 numbers is zero, them at most, how many may be greater than zero?
(a) 0
(b) 1
(c) 10
(d) 19
Answer: (d) 19
Explanation :
Mean of 20 numbers = 0
Hence, sum of 20 numbers = 0 × 20 = 0
Now, the mean can be zero if
Sum of 10 numbers is (S) and the sum of remaining 10 numbers is (– S),
Sum of 11 numbers is (S) and the sum of remaining 9 numbers is (– S),

Question. For a symmetrical frequency distribution, we have :
(a) Mean < Mode < Median
(b) Mean < Mode > Median
(c) Mean = Mode = Medain
(d) Mode = 12 + 12 (Mean + Medain)
Answer: (c) Mean = Mode = Medain
Explanation :
For a symmetrical distribution, we have \( \text{Mean} = \text{Mode} = \text{Median} \)

Question. The median and mode of a frequency distribution are 26 and 29 respectively. Then, the mean is :
(a) 27.5
(b) 24.5
(c) 28.4
(d) 25.8
Answer: (b) 24.5
Explanation :
\( \text{Median} = 26 \)
\( \text{Mode} = 29 \)
\( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
Hence, \( \text{Mean} = \frac{3 \text{ Median} – \text{Mode}}{2} \)
\( = \frac{3(26) – 29}{2} = \frac{78 – 29}{2} = \frac{49}{2} = 24.5 \)

Question. The algebraic sum of the deviations of a frequency distribution from its mean is :
(a) Always positive
(b) Always negative
(c) 0
(d) A non-zero number
Answer: (c) 0
Explanation :
The algebraic sum of the deviations of a frequency distribution from its mean is zero.
Let \( x_1, x_2, x_3, .... x_n \) are observations and \( \bar{x} \) is the mean
\( \sum(x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + (x_3 - \bar{x}) + .... + (x_n - \bar{x}) \)
\( = (x_1 + x_2 + x_3 + .... + x_n) - n\bar{x} \)
\( = n\bar{x} - n\bar{x} = 0 \)

Question. If the mean of a data is 27 and its median is 33. Then, the mode is :
(a) 30
(b) 43
(c) 45
(d) 47
Answer: (c) 45
Explanation :
\( \text{Mean} = 27 \)
\( \text{Median} = 33 \)
\( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
\( = 3 \times 33 – 2 \times 27 \)
\( = 99 – 54 = 45 \)

Question. If the median of the data 4, 7, x – 1, x – 3, 16, 25 written is ascending order is 13, then x is equal to :
(a) 13
(b) 14
(c) 15
(d) 16
Answer: (c) 15
Explanation :
Data is ascending order: 4, 7, x – 1, x – 3, 16, 25
\( N = 6 \text{(even)} \)
\( \text{Median} = \frac{(\frac{6}{2})^{\text{th}} \text{ value} + (\frac{6}{2}+1)^{\text{th}} \text{ value}}{2} \)
\( = \frac{(x-1) + (x-3)}{2} = \frac{2x - 4}{2} = x - 2 \)
\( x – 2 = 13 \)
\( x = 15 \)

Question. If mode of a series exceeds its mean by 12, then mode exceeds the median by :
(a) 4
(b) 8
(c) 6
(d) 10
Answer: (b) 8
Explanation :
Given : \( \text{Mode} – \text{Mean} = 12 \)
We know that \( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
\( \text{Mode} – \text{Mean} = 3(\text{Median} – \text{Mean}) \)
\( 12 = 3(\text{Median} – \text{Mean}) \)
\( \text{Median} – \text{Mean} = 4 \)...(i)
Again,
\( \text{Mode} = 3 \text{ Median} – 2 \text{ Mean} \)
\( 2 \text{ Mode} = 6 \text{ Median} – 4 \text{ Mean} \)
\( \text{Mode} – \text{Mean} + \text{Mode} = 6 \text{ Median} – 5 \text{ Mean} \)
\( 12 + (\text{Mode} – \text{Median}) = 5 (\text{Median} – \text{Mean}) \)
\( 12 + (\text{Mode} – \text{Median}) = 5(4) \) [Using (i)]
\( \text{Mode} – \text{Median} = 20 – 12 = 8 \)