CBSE Class 10 Mathematics Statistics Assignment Set 21

Exercise-1 Objective Questions: Introduction and Arithmetic Mean

Question A-1. If the arithmetic mean of 5, 7, 9, x is 9 then the value of x is :
(A) 11   (B) 15   (C) 18   (D) 16
Answer:
Ans. (B) 15
5 + 7 + 9 + x = 9 × 4 = 36, so x = 15.

Teacher's Note:
1. Sum = mean × number of observations.

Question A-2. If the first five elements of the set x₁, x₂, …, x₁₀ are replaced by xᵢ + 5, i = 1, 2, 3, 4, 5 and the next five elements are replaced by xⱼ − 5, j = 6, 7, …, 10, then the mean will change by :
(A) 0   (B) (n + 1)/2   (C) 10   (D) 25
Answer:
Ans. (A) 0
The total change in the sum = 5 × (+5) + 5 × (−5) = 0.
So the mean changes by 0.

Teacher's Note:
1. Equal additions and subtractions cancel in the sum.

Question A-3. Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, then find the value of x.
(A) 45   (B) 55   (C) 35   (D) 40
Answer:
Ans. (A) 45
Total = 50 × 4 = 200. 36 + 44 + 75 + x = 200, so x = 45.

Teacher's Note:
1. Total marks = average × number of subjects.

Question A-4. If uᵢ = (xᵢ − 20)/10, Σfᵢuᵢ = 30 and Σfᵢ = 40, then find the value of x̄.
(A) 30   (B) 25   (C) 28.5   (D) 27.5
Answer:
Ans. (D) 27.5
x̄ = a + h (Σfᵢuᵢ / Σfᵢ) with a = 20 and h = 10.
x̄ = 20 + 10 × \( \dfrac{30}{40} \) = 27.5.

Teacher's Note:
1. Read a and h from the definition of uᵢ.

Question A-5. The mean of 20 numbers is 13. The new mean if each observation is increased by 5, is
(A) 13   (B) 18   (C) 65   (D) 8
Answer:
Ans. (B) 18
Adding a constant to every observation adds it to the mean: 13 + 5 = 18.

Teacher's Note:
1. Adding a constant shifts the mean by that constant.

Median

Question B-1. The median of the following series is 520, 20, 340, 190, 35, 800, 1210, 50, 80.
(A) 1210   (B) 520   (C) 190   (D) 35
Answer:
Ans. (C) 190
Arranged in ascending order: 20, 35, 50, 80, 190, 340, 520, 800, 1210 (9 values).
Median = 5th value = 190.

Teacher's Note:
1. Always arrange in order before choosing the middle value.

Question B-3. A data has 25 observations (arranged in descending order). Which observation represents the median?
(A) 12th   (B) 13th   (C) 14th   (D) 15th
Answer:
Ans. (B) 13th
For n = 25, median = \( \left( \dfrac{25 + 1}{2} \right) \)th = 13th observation.

Teacher's Note:
1. For odd n, the median is the ((n + 1)/2)th observation.

Question B-4. The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
(A) 20   (B) 21   (C) 22   (D) 23
Answer:
Ans. (B) 21
There are 10 observations, so the median is the mean of the 5th and 6th: \( \dfrac{(x + 2) + (x + 4)}{2} = x + 3 = 24 \).
x = 21.

Teacher's Note:
1. For even n, average the two middle observations.

Question B-5. If in a data, 10 numbers are arranged in increasing order, and the 7th entry is increased by 4, then the median increases by -
(A) zero   (B) 4   (C) 6   (D) 5
Answer:
Ans. (A) zero
The median is the mean of the 5th and 6th entries. Increasing the 7th entry does not change them (the order is preserved).
The median increases by zero.

Teacher's Note:
1. Only the middle entries decide the median.

Mode

Question C-1. The mode of the distribution 3, 5, 7, 4, 2, 1, 4, 3, 4 is :
(A) 7   (B) 4   (C) 3   (D) 1
Answer:
Ans. (B) 4
4 occurs three times, 3 occurs twice and the rest once, so the mode is 4.

Teacher's Note:
1. The mode is the most frequent value.

Question C-2. If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be :
(A) 7.2   (B) 8.2   (C) 9.2   (D) 10.2
Answer:
Ans. (C) 9.2
Mode = 3 Median − 2 Mean = 27 − 17.8 = 9.2.

Teacher's Note:
1. Use the empirical relation.

Question C-3. If mode = 80 and mean = 110, then the median is :
(A) 110   (B) 120   (C) 100   (D) 90
Answer:
Ans. (C) 100
Mode = 3 Median − 2 Mean ⇒ 80 = 3 Median − 220 ⇒ Median = 100.

Teacher's Note:
1. Rearrange the empirical relation for the median.

Question C-4. If the mode of the following data is 7, then the value of k in 2, 4, 6, 7, 5, 6, 10, 6, 7, 2k + 1, 9, 7, 13 is :
(A) 3   (B) 7   (C) 4   (D) 2
Answer:
Ans. (A) 3
Currently 6 occurs three times and 7 occurs three times, so for 7 to be the mode, 2k + 1 must be 7.
2k + 1 = 7 gives k = 3.

Teacher's Note:
1. The extra entry must break the tie in favour of 7.

Question C-5. For a symmetrical distribution, which is correct :
(A) Mean > Mode > Median   (B) Mean < Mode < Median   (C) Mode = (Mean − Median)/2   (D) Mean = Median = Mode
Answer:
Ans. (D) Mean = Median = Mode
In a symmetrical distribution the three measures coincide: Mean = Median = Mode.

Teacher's Note:
1. Symmetry puts the mean, median and mode at the same value.

Class mark and cumulative frequency

Question D-3. The classmark of a class interval is :
(A) Lower limit + Upper limit   (B) Upper limit − Lower limit   (C) (1/2) (Lower limit + Upper limit)   (D) (1/4) (Lower limit + Upper limit)
Answer:
Ans. (C) (1/2) (Lower limit + Upper limit)
The class mark is the mid-point: ½ (lower limit + upper limit).

Teacher's Note:
1. Class mark = mid-value of the class.

Question D-5. Which of the following is not a measure of central tendency?
(A) Mean   (B) Median   (C) Range   (D) Mode
Answer:
Ans. (C) Range
Mean, median and mode describe the centre; range measures the spread.

Teacher's Note:
1. Range = largest value − smallest value.