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Test: Statistics - Class 10 MCQ


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15 Questions MCQ Test The Complete SAT Course - Test: Statistics

Test: Statistics for Class 10 2024 is part of The Complete SAT Course preparation. The Test: Statistics questions and answers have been prepared according to the Class 10 exam syllabus.The Test: Statistics MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Statistics below.
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Test: Statistics - Question 1

Comprehension:

Direction: Based on the following information, answer the questions.

For an experiment, a sample of 25 observations from normal distribution with mean 98.6 and s.d. 17.2 was selected. Answer the following questions:
What is P(92 <  < 102)?

Detailed Solution for Test: Statistics - Question 1

Concept:

If a random sample of size n is taken from a population with mean μ and s.d. σ, then the sample distribution of X (sample mean) has a mean equal to population mean i.e., μ and a s.d. σM
Therefore,  follows standard normal distribution.

Calculation:

We have μ = 98.6, σM=2.86

⇒ - 1.918 < Z < 0.988

= P(- 1.918 < Z < 0.988)

= FZ(0.988) − FZ(−1.918)

= 0.3389 + 0.4726

= 0.8115

Test: Statistics - Question 2

Comprehension:

Direction: Based on the following information, answer the questions.

For an experiment, a sample of 25 observations from normal distribution with mean 98.6 and s.d. 17.2 was selected. Answer the following questions:

What is the standard error of mean, for the population size to be 100?

Detailed Solution for Test: Statistics - Question 2

Concept:

The standard error of mean:

Calculation:

We have σ = 17.2

Sample size, n = 25

Population size = 100

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Test: Statistics - Question 3

What is the mean of the range, mode and median of the data given below?

5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Detailed Solution for Test: Statistics - Question 3

Given:

The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:

The mode is the value that appears most frequently in a data set

At the time of finding Median

First, arrange the given data in the ascending order and then find the term

Formula used:

Mean = Sum of all the terms/Total number of terms

Median = {(n + 1)/2}th term when n is odd 

Median = 1/2[(n/2)th term + {(n/2) + 1}th] term when n is even

Range = Maximum value – Minimum value 

Calculation:

Arranging the given data in ascending order 

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19

Here, Most frequent data is 4 so 

Mode = 4

Total terms in the given data, (n) = 15 (It is odd)

Median = {(n + 1)/2}th term when n is odd 

⇒ {(15 + 1)/2}th term 

⇒ (8)th term

⇒ 6 

Now, Range = Maximum value – Minimum value 

⇒ 19 – 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3

⇒ (17 + 4 + 6)/3 

⇒ 27/3 = 9

∴ The mean of the Range, Mode and Median is 9

Test: Statistics - Question 4

If mean and mode of some data are 4 & 10 respectively, its median will be:

Detailed Solution for Test: Statistics - Question 4

Concept:

Mean: The mean or average of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set.

Mode: The mode is the value that appears most frequently in a data set.

Median: The median is a numeric value that separates the higher half of a set from the lower half. 

Relation b/w mean, mode and median:

Mode = 3(Median) - 2(Mean)

Calculation:

Given that,

mean of data = 4 and mode of  data = 10

We know that

Mode = 3(Median) - 2(Mean)

⇒ 10 = 3(median) - 2(4)

⇒ 3(median) = 18

⇒ median = 6

Hence, the median of data will be 6.

Test: Statistics - Question 5

Find the median of the given set of numbers 2, 6, 6, 8, 4, 2, 7, 9

Detailed Solution for Test: Statistics - Question 5

Concept:

Median: The median is the middle number in a sorted- ascending or descending list of numbers.

Case 1: If the number of observations (n) is even

Case 2: If the number of observations (n) is odd

Calculation:

Given values 2, 6, 6, 8, 4, 2, 7, 9

Arrange the observations in ascending order:

2, 2, 4, 6, 6, 7, 8, 9

Here, n = 8 = even

As we know, If n is even then,


Hence Median = 6

Test: Statistics - Question 6

The mean of four numbers is 37. The mean of the smallest three of them is 34. If the range of the data is 15, what is the mean of the largest three?

Detailed Solution for Test: Statistics - Question 6

Calculation:

Let the numbers be x1, x2, x3, x4.

The mean of four numbers x1, x2, x3, x4 = 37

The sum of four numbers x1, x2, x3, x= 37 × 4 = 148.

The mean of the smallest three numbers x1, x2, x3 = 34

The sum of the smallest three numbers x1, x2, x3 = 34 × 3 = 102.

∴ The value of the largest number x= 148 – 102 = 46.

The Range (Difference between largest and smallest value) x4 – x1 = 15.

∴ Smallest number x1 = 46 – 15 = 31.

Now,

The sum of x2, x= Total sum – (sum of smallest and largest number).

⇒ 148 – (46 + 31)

⇒ 148 – 77

⇒ 71

Now,

The mean of the Largest three numbers x2, x3, x= (71 + 46)/3 = 117/3 = 39

Test: Statistics - Question 7

The data given below shows the marks obtained by various students.

What is the mean marks (Correct up to two decimal places) of given data? 

Detailed Solution for Test: Statistics - Question 7

⇒ n = total frequency

∑fx = Sumoftheproductofmid − intervalvaluesandtheircorrespondingfrequency

Mid value of 10 – 12 = (10 + 12)/2 = 11

Mid value of 12 – 14 = (12 + 14 )/2 = 13

Mid value of 14 – 16 = (14 + 16 )/2 = 15

Mid value of 16 – 18 = (16 + 18 )/2 = 17

Mid value of 18 – 20 = (18 + 20 )/2 = 19

⇒ Mean = 14.67

∴The mean marks of the given data are 14.67

Test: Statistics - Question 8

Comprehension:

Direction: Based on the following information, answer the questions.

For an experiment, a sample of 25 observations from normal distribution with mean 98.6 and s.d. 17.2 was selected. Answer the following questions:

Find the corresponding probability given a sample of 36

Detailed Solution for Test: Statistics - Question 8

Concept:

If a random sample of size n is taken from a population with mean μ and s.d. σ, then the sample distribution of X (sample mean) has a mean equal to population mean i.e., μ and a s.d. σM
Therefore,  follows standard normal distribution.

Calculation:

We have μ = 98.6, σM=2.86

⇒ - 2.3 < Z < 1.19

= P(- 2.3 < Z < 1.19)

= FZ(1.19) − FZ(−2.3)

= 0.4843 + 0.3830

= 0.8723

Test: Statistics - Question 9

Comprehension:

Direction: Based on the following information, answer the questions.

For an experiment, a sample of 25 observations from normal distribution with mean 98.6 and s.d. 17.2 was selected. Answer the following questions:

What is the standard error of mean for a sample size n = 36?

Detailed Solution for Test: Statistics - Question 9

Concept:

The standard error of mean:

Calculation:

We have σ = 17.2

Sample size, n = 36

Test: Statistics - Question 10

Comprehension:

Direction: Based on the following information, answer the questions.

For an experiment, a sample of 25 observations from normal distribution with mean 98.6 and s.d. 17.2 was selected. Answer the following questions:

What is the standard error of mean?

Detailed Solution for Test: Statistics - Question 10

Concept:

The standard error of mean:

Calculation:

We have σ = 17.2

Sample size, n = 25

Test: Statistics - Question 11

Find the mean of given data:

Detailed Solution for Test: Statistics - Question 11

Formula used:

The mean of grouped data is given by,

Xi = mean of ith class

fi = frequency corresponding to ith class

Given:

Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

Then,

We know that, mean of grouped data is given by

= 35.7

Hence, the mean of the grouped data is 35.7

Test: Statistics - Question 12

If the standard deviation of 0, 1, 2, 3 ______ 9 is K, then the standard deviation of 10, 11, 12, 13 _____ 19 will be:

Detailed Solution for Test: Statistics - Question 12

Formula Used∶

  • σ2 = ∑(xi – x)2/n
  • Standard deviation is same when each element is increased by the same constant

Calculation:

Since each data increases by 10,

There will be no change in standard deviation because (xi – x) remains same.

∴ The standard deviation of 10, 11, 12, 13 _____ 19 will be will be K.

Test: Statistics - Question 13

What is the standard deviation of the observations

−√6, −√5, −√4, −1, 1, √4, √5, √6?

Detailed Solution for Test: Statistics - Question 13

Concept:

Standard deviation:

The standard deviation of the observation set {xi, i = 1, 2, 3,⋯} is given as follows:

Where N = size of the observation set and μ = mean of the observations.
Calculations:
First, we will calculate the mean of the given observations.

Therefore, the numerator inside the square root term of the standard deviation formula will simply be equal to (xi − μ)2 = xi2.

Now we observe that N = 8.

Therefore, the standard deviation is given as follows:


Therefore, the standard deviation of the given observations is 2.

Test: Statistics - Question 14

If the mean of a frequency distribution is 100 and the coefficient of variation is 45%, then what is the value of the variance?

Detailed Solution for Test: Statistics - Question 14

Concept:

Coefficient of variation = Standard Deviation/Mean

Variance = (Standard Deviation)2

Calculation:

Given coefficient of variation = 45% = 0.45

And mean = 100

As Coefficient of variation = Standard Deviation/Mean

0.45 = Standard Deviation/100

Standard Deviation = 100 × 0.45

SD = 45

∴ Variance = 452 = 2025

Test: Statistics - Question 15

Consider the following distribution:

The frequency of class interval 30,000 - 40,000 is:

Detailed Solution for Test: Statistics - Question 15

Concept:

  • Frequency of any class = cumulative frequency of class - cumulative frequency of preceding class
  • Cumulative frequency is the sum of all the previous frequencies up to the current point.

Calculation:

We are already having the cumulative frequencies. Let's find out the frequencies:

Hence, the frequency of class intervals 30,000 - 40,000 is 6.

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