CBSE Class 10 Statistics Sure Shot Questions Set I

Question. If the value of mean and mode are respectively 30 and 15, then Median =
(a) 22.5
(b) 24.5
(c) 25
(d) 26
Answer: (c) 25

Question. While computing mean of grouped data, we assume that the frequencies are
(a) evenly distributed over all the class
(b) centered at the class marks of the class
(c) centered the upper limits of the class
(d) centered the lower limits of the class
Answer: (b) centered at the class marks of the class

Question. The wickets taken by a bowler in 10 cricket matches are as follows: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. Find the mode of the data.
(a) 2
(b) 4
(c) 3
(d) 1
Answer: (a) 2

Question. If mean of the distribution is 7.5, then p=
X: 3, 5, 7, 9, 11, 13
F: 6, 8, 15, P, 8, 4
(a) 2
(b) 4
(c) 3
(d) 6
Answer: (c) 3

Question. If the arithmetic Mean of \( x, x + 3, x + 6, x + 9 \) and \( x + 12 \) is 10, then \( x \) =?
(a) 1
(b) 2
(c) 6
(d) 4
Answer: (d) 4

Question. If the sum of frequencies is 24, then the value of \( x \) in the observation: \( x, 5, 6, 1, 2, \) will be
(a) 4
(b) 6
(c) 8
(d) 10
Answer: (d) 10

Question. In the formula \( \bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right) \), for finding the mean of grouped frequency distribution, \( u_i \) =
(a) \( \frac{x_i+a}{h} \)
(b) \( h (x_i – a) \)
(c) \( \frac{x_i – a}{h} \)
(d) \( \frac{a – x_i}{h} \)
Answer: (c) \( \frac{x_i – a}{h} \)

Question. Consider the following frequency distribution:
0 – 5 | 6 – 11 | 12-17 | 18-23 | 24-29
13 | 10 | 15 | 8 | 11
The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5
Answer: (b) 17.5

Question. The algebraic sum of the deviations of a frequency distribution from its mean is always,
(a) greater than zero
(b) less than zero
(c) zero
(d) a non-zero number
Answer: (c) zero

Question. The empirical relationship between the three measures of central tendency is
(a) 3 Median = Mode + 2 Mean
(b) 2 Median = Mode + 2 Mean
(c) 3 Median = Mode + Mean
(d) 3 Median = Mode – 2 Mean
Answer: (a) 3 Median = Mode + 2 Mean

Question. The median of the data 13, 15, 16, 17, 19, 20 is:
(a) 30/2
(b) 31/2
(c) 33/2
(d) 35/2
Answer: (c) 33/2

Question. Mode and mean of a data are 12k and 15k. Median of the data is
(a) 12k
(b) 14k
(c) 15k
(d) 16k
Answer: (b) 14k

Question. Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, and 80 were wrongly read as 40, 20, and 50 respectively. The correct mean is
(a) 48
(b) 49
(c) 50
(d) 60
Answer: (c) 50

Question. Mode is the
(a) middle most frequent value
(b) least frequent value
(c) maximum frequent value
(d) None of the options
Answer: (c) maximum frequent value

Question. If the mean of frequency distribution is 7.5 and \( \sum f_i x_i = 120 + 3k \), \( \sum f_i = 30 \), then k is equal to:
(a) 40
(b) 35
(c) 50
(d) 45
Answer: (b) 35

Assertion and Reason Questions

Question. Statement A (Assertion): The mode of the call received on 7 consecutive day 11, 13, 13, 17, 19, 23, 25 is 13.
Statement R (Reason): Mode is the value that appears most frequent;
(a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) if both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) if Assertion (A) is true but reason (R) is false.
(d) if Assertion (A) is false but reason (R) is true.
Answer: (a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Statement A (Assertion): The runs scored by a batsman in 5 ODIs are 31, 97, 112, 63, and 12. The standard deviation is 25.79.
Statement R (Reason): mean = total sum of number in data sets. / Total number in data sets.
(a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) if both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) if Assertion (A) is true but reason (R) is false.
(d) if Assertion (A) is false but reason (R) is true.
Answer: (b) if both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

Question. Statement A (Assertion): If the median of the given data 26, 29, 42, 53, x, x + 2, 70, 75, 82, 93, is 65 then the value of x is 64.
Statement R (Reason): When the number of observations (n) is odd the median is the value of the \( (n+1)/2 \) th observation.
(a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) if both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) if Assertion (A) is true but reason (R) is false.
(d) if Assertion (A) is false but reason (R) is true.
Answer: (a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Statement A (Assertion): Median of 51, 70, 65, 82, 60, 68, 62, 95, 55, 64, 58, 75, 80, 85, and 90 is 68.
Statement R (Reason): When n observations are arranged in an ascending order and n is odd, then median = value of \( 1/2(n+1) \) th observation.
(a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) if both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) if Assertion (A) is true but reason (R) is false.
(d) if Assertion (A) is false but reason (R) is true.
Answer: (a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Statement A (Assertion): If mean & median of an asymmetrical distribution are 58 & 61 respectively, then Mode = 67.
Statement R (Reason): For an asymmetrical distribution Mode = 3 Median − 2 Mean.
(a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) if both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) if Assertion (A) is true but reason (R) is false.
(d) if Assertion (A) is false but reason (R) is true.
Answer: (a) if both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

SHORT ANSWER TYPE QUESTIONS 

Question. If the mean of 4 numbers, 2, 6, 7 and a is 15 and also the mean of other 5 numbers, 6, 18, 1, a, b is 50. What is the value of b?
Answer: 180

Question. A student scored the following marks in 6 subjects: 30, 19, 25, 30, 27, 30. Find his modal score.
Answer: 30

Question. The modal class of the grouped size frequency table given below is
5-5.2 | 5.2-5.4 | 5.4-5.6 | 5.6-5.8 | 5.8-6.0
20 | 5 | 5 | 12 | 8
Answer: 5-5.2

Question. The points scored by a basketball team in a series of matches are as follows: 17, 2, 7, 27, 25, 5, 14, 18, 10. Find the median.
Answer: 14

Question. Find the mean of the following distribution:
X: 5, 10, 15, 20, 25
F: 4, 12, 20, 28, 6
Answer: 15.1

Question. Find the mean of first five Prime numbers.
Answer: 5.6

Question. If the mode of a distribution is 18 and its mean is 12, then find its median.
Answer: 14

Question. Write the empirical relationship between the three measures of central tendency.
Answer: Mode = 3 Median – 2 Mean

Question. A data has 15 observations arranged in descending order. Which observation represents the Median of the data?
Answer: 8th term

Question. Find the class size of the given class intervals.
Class Interval: 0 – 6, 6 – 12, 12 – 18, 18 – 24, 24 – 30, 30 – 46
Answer: 6 (Note: Last class size is 16, others are 6)