Description

This mock test of Test: Statistics- 1 for Class 9 helps you for every Class 9 entrance exam.
This contains 25 Multiple Choice Questions for Class 9 Test: Statistics- 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Statistics- 1 quiz give you a good mix of easy questions and tough questions. Class 9
students definitely take this Test: Statistics- 1 exercise for a better result in the exam. You can find other Test: Statistics- 1 extra questions,
long questions & short questions for Class 9 on EduRev as well by searching above.

QUESTION: 1

A student collects information about the number of school going children in a locality consisting of a hundred households. The data collected by him is

Solution:

QUESTION: 2

Class mark of a particular class is 9.5 and the class size is 6, then the class interval is

Solution:

QUESTION: 3

One of the sides of a frequency polygon is

Solution:

QUESTION: 4

Out of sixteen observations arranged in an ascending order, the 8^{th} and 9^{th} observations are 25 and 27. Then, the median is

Solution:

QUESTION: 5

Vihaan has marks of 92, 85, and 78 in three mathematics tests. In order to have an average of exactly 87 for the four math tests, he should obtain

Solution:

QUESTION: 6

To analyse the election results, the data is collected from a newspapers. The data thus collected is known as

Solution:

QUESTION: 7

The class size of a distribution is 25 and the first class interval is 200-224. Then, the class marks of first two class intervals are

Solution:

class mark of 1st interval=212

class mark of 2nd interval=236.5

Step-by-step explanation:

First Class interval=200-224

class mark=upper limit +lower limit divided by 2..

class mark of first interval is =200+224 divided by 2=212

Second class interval=224-249

class mark of second interval =224 +249 divided by 2

answer= 236.5

QUESTION: 8

Which of the following is NOT a common measure of central tendency?

Solution:

QUESTION: 9

What is the median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84?

Solution:

QUESTION: 10

If the mean of x and 1/x is M, then the mean of x^{2} and

Solution:

If the mean of x and 1/2 is M. then the mean of

x² and 1/x² is

Mean of x and 1/x is M. So

( x +1/x)/2=M

Squaring on both sides,

(x +1/x)²/2²=M²

(x² + 1/x²+2*x*1/x)/4=M²

x²+1/x²+2=4 m²

x²+ 1/x²=4m²-2

Dividing by 2 on both sides,

(x²+1/x²)/2=(4m²-2)/2

= (x²+1/x²)/2 =2 *(2m²-1)/2=2m²-1

=Mean of x² and 1/x²=(2m²-1)

QUESTION: 11

Which of the following variables are discrete ? 1. Size of shoes, 2. Number of pages in a book, 3. Distance travelled by a train, 4. Time

Solution:

QUESTION: 12

In a frequency distribution, the mid-value of a class is 60.5 and the width of the class is 10. The lower limit of the class is

Solution:

QUESTION: 13

The mean for counting numbers through 100 is

Solution:

QUESTION: 14

Mode of a set of observations is the value which

Solution:

QUESTION: 15

The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers becomes 20. The excluded number is

Solution:

QUESTION: 16

For a given data, the difference between the maximum and minimum observation is known as its

Solution:

QUESTION: 17

The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. The upper class limit of the highest class is

Solution:

Let x and y be the upper and lower class limit of frequency distribution.

Given, width of the class = 5

⇒ x-y= 5 …(i)

Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get

x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15.

Hence, the upper class limit of the highest class

=(Number of continuous classes x Class width + Lower class limit of the lowest class)

= 5 x 5+10 = 25+10=35

Hence,’the upper class limit of the highest class is 35.

Alternate Method

After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.

Thus, the highest class is 30-35,

Hence, the upper limit of this class is 35.

QUESTION: 18

The mean of first four prime numbers is

Solution:

QUESTION: 19

The mode of 4, 6, 7, 8, 12, 11, 13, 9, 13, 9, 7, 8, 9 is

Solution:

QUESTION: 20

The mean of five observations is 15. If the mean of first three observations is 14 and that of last three is 17, then the third observation is

Solution:

⇒ It is given that Mean of 5 observations is 15.

∴ The sum of observations =15×5=75.

∴ It is given that mean of the first 3 observations is 14.

∴ The sum of first three observations =14×3=42.

⇒ Given that mean of the last 3 observations is 17.

∴ The sum of the last three observations =17×3=51

∴ The third observation =(42+51)−75

=93−75

=18

QUESTION: 21

In an examination, ten students scored the following marks: 60, 58, 90, 51, 47, 81, 70, 95, 87, 99. The range of this data is

Solution:

QUESTION: 22

In a bar graph, 0.25 cm length of a bar represents 100 people. Then, the length of bar which represents 2000 people is

Solution:

QUESTION: 23

If, for the set of observations 4, 7, x, 8, 9, 10 the mean is 8, then x is equal to

Solution:

QUESTION: 24

A set of data consists of six numbers: 7, 8, 8, 9, 9 and x. The difference between the modes when x = 9 and x = 8 is

Solution:
It is not the matter of value of x .since modes are 9 and 8 .therefore difference will remain same i.e 1

QUESTION: 25

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. The mean of the given number is

Solution:

Total numbers =50

Mean of numbers after subtracting 53 from each =3.5

Sum of numbers after subtracting 53 from each =3.5×50=175

Sum of the original numbers =175+53×50=2825

Mean of the original numbers =2825/50=56.5

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