Class 9 Maths: Statistics – Extra Questions with Answers
Q. 1 What is the median of the numbers 5, 7, 9, 12, 13, 17, 18?
Check Solution
Ans: C
Solution:
Arrange the numbers in ascending order: 5, 7, 9, 12, 13, 17, 18.
The median is the middle number. There are 7 numbers, so the middle number is the 4th one.
Answer: 12
Q. 2 What is the range of the data set: 15, 20, 6, 5, 30, 35, 92, 35, 90, 18, 81?
Check Solution
Ans: B
Solution:
To find the range, subtract the smallest value from the largest value. The smallest value is 5, and the largest value is 92.
Range = 92 – 5 = 87
Answer: 87
Q. 3 What is the mode of the following distribution? | Marks | 10 | 20 | 30 | 33 | 40 | 45 | | :—————- | :- | :- | :- | :- | :- | :- | | Number of students | 2 | 4 | 8 | 10 | 1 | 3 |
Check Solution
Ans: A
Solution: The mode is the value that appears most frequently. In this distribution, the mark 33 appears most frequently (10 times).
Answer: 33
Q. 4 The class marks of a frequency distribution are 10, 15, 20, 25, ….. If the class width is 5, what is the lower class limit of the class corresponding to the class mark 20?
Check Solution
Ans: C
Solution:
Class mark = (Lower limit + Upper limit) / 2.
Class width = Upper limit – Lower limit = 5
Lower limit = Class mark – (Class width / 2)
Lower limit = 20 – (5 / 2) = 20 – 2.5 = 17.5
Answer: 17.5
Q. 5 A frequency distribution of test scores has the following values: Scores (x): 60, 70, 80, 90, 100. Frequency (f): 4, 6, 7, 2, 1. Calculate the mean test score.
Check Solution
Ans: D
Solution:
Mean = (Σxf) / Σf = [(60*4)+(70*6)+(80*7)+(90*2)+(100*1)] / (4+6+7+2+1) = (240+420+560+180+100) / 20 = 1500 / 20 = 75
Answer: 75
Q. 6 If the mean of 1000, 1002, 1004, 1006, 1008 is $n^3 + 4$, then find the value of n.
Check Solution
Ans: A
Solution:
The mean of the numbers is (1000 + 1002 + 1004 + 1006 + 1008)/5 = 5020/5 = 1004.
We are given that the mean is n^3 + 4.
Therefore, n^3 + 4 = 1004.
n^3 = 1000.
n = 10.
Answer: 10
Q. 7 Find the mode of the following: 42, 40, 49, 53, 47, 63, 55, 41, 97, 53, 99, 41, 43, 53, 61, 41.
Check Solution
Ans: B
Solution:
Count the occurrences of each number:
41: 3
40: 1
42: 1
49: 1
53: 3
47: 1
63: 1
55: 1
97: 1
99: 1
43: 1
61: 1
The mode is the number that appears most frequently. Both 41 and 53 appear 3 times.
Answer: 41, 53
Q. 8 The following observations have been arranged in ascending order. If the median of the data is 65, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Check Solution
Ans: A
Solution:
Since there are 10 observations, the median is the average of the 5th and 6th values. The 5th value is x and the 6th is x+2. So (x + x + 2)/2 = 65.
2x + 2 = 130
2x = 128
x = 64
Answer: 64
Q. 9 The mid-value of a class interval is 45, and the lower limit is 40. Find the upper limit.
Check Solution
Ans: A
Solution:
Mid-value = (Lower limit + Upper limit) / 2
45 = (40 + Upper limit) / 2
90 = 40 + Upper limit
Upper limit = 50
Answer: 50
Q. 10 Let $M$ be the midpoint of a class in a frequency distribution and $W$ be the class width. If the upper class boundary is $U$, what is the lower class boundary of the class?
Check Solution
Ans: A
Solution:
Since M is the midpoint, $M = \frac{U + L}{2}$, where L is the lower class boundary. We also know $W = U – L$. We can rearrange the midpoint formula to find the lower class boundary: $L = 2M – U$. Since $W = U – L$, then $L = U – W$.
Answer: $U – W$
Next Topic: Number Systems
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