Statistics - Free MCQ Practice Test with solutions, Class 10 The Complete

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. σ_M
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)

= F_Z(0.988) − F_Z(−1.918)

= 0.3389 + 0.4726

= 0.8115

Concept:

The standard error of mean:

Calculation:

We have σ = 17.2

Sample size, n = 25

Population size = 100

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

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.

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

Calculation:

Let the numbers be x_1, x₂, x₃, x₄.

The mean of four numbers x_1, x₂, x₃, x₄ = 37

The sum of four numbers x_1, x₂, x₃, x₄ = 37 × 4 = 148.

The mean of the smallest three numbers x_1, x₂, x₃ = 34

The sum of the smallest three numbers x_1, x₂, x₃ = 34 × 3 = 102.

∴ The value of the largest number x₄ = 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, x₃ = Total sum – (sum of smallest and largest number).

⇒ 148 – (46 + 31)

⇒ 148 – 77

⇒ 71

Now,

The mean of the Largest three numbers x₂, x₃, x₄ = (71 + 46)/3 = 117/3 = 39

⇒ 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

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)

= F_Z(1.19) − F_Z(−2.3)

= 0.4843 + 0.3830

= 0.8723

Concept:

The standard error of mean:

Calculation:

We have σ = 17.2

Sample size, n = 36

Concept:

The standard error of mean:

Calculation:

We have σ = 17.2

Sample size, n = 25

Formula used:

The mean of grouped data is given by,

X_i = mean of i^th class

f_i = frequency corresponding to i^th class

Given:

Calculation:

Now, to calculate the mean of data will have to find ∑f_iX_i and ∑f_i 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

Formula Used∶

• σ² = ∑(x_i – 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 (x_i – x) remains same.

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

Concept:

Standard deviation:

The standard deviation of the observation set {x_i, 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 (x_i − μ)² = x_i².

Now we observe that N = 8.

Therefore, the standard deviation is given as follows:


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

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

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.