Standard Deviation MCQ Quiz - Objective Question with Answer for Standard Deviation - Download Free PDF

Given:

The weights (in kg) of 10 animals are 75, 77, 77, 79, 79, 81, 81, 83, 85, and 85. 

Concept Used:

Steps to find the standard deviation,

1. Calculate the mean (average) of the set of numbers.
2. For each number in the set, subtract the mean and square the result.
3. Find the average of the squared differences calculated in step 2.
4. Take the square root of the result from step 3.

Calculation:

First, find the mean of the weights,

⇒ Mean = (75+77+77+79+79+81+81+83+85+85)/10 = 80.2

Next, find the deviations from the mean for each weight:

⇒ (75 - 80.2), (77 - 80.2), (77 - 80.2), (79 - 80.2), (79 - 80.2), (81 - 80.2), (81 - 80.2), (83 - 80.2), (85 - 80.2), (85 - 80.2)

⇒ -5.2, -3.2, -3.2, -1.2, -1.2, 0.8, 0.8, 2.8, 4.8, 4.8

Then, square each deviation and add them up:

⇒ 27.04 + 10.24 + 10.24 + 1.44 + 1.44 + 0.64 + 0.64 + 7.84 + 23.04 + 23.04

⇒ 105.6

Divide the sum by the number of weights (10) and take the square root:

⇒ √10.56 = 3.25

∴ The standard deviation of the weights of the animals is 3.25 kg.

Given:

Coefficient of variation of two distributions = 75 and 80

Standard deviations of two distributions = 15 and 16

Formula used:

Coefficient of Variation (CV) = (Standard Deviation (σ) / Arithmetic Mean (μ)) × 100

Calculation:

For the first distribution:

CV = 75, σ = 15

⇒ 75 = (15 / μ) × 100

⇒ μ = (15 × 100) / 75

⇒ μ = 20

For the second distribution:

CV = 80, σ = 16

⇒ 80 = (16 / μ) × 100

⇒ μ = (16 × 100) / 80

⇒ μ = 20

∴ The arithmetic means of the two distributions are 20 and 20, respectively

Given:

Variance of the data = 361

Formula used:

Standard Deviation (SD) = √(Variance)

Calculation:

SD = √361

⇒ SD = 19

∴ The correct answer is option (3).

Concept:

Mean \( \mu(A) \):

\(\mu(A) = \frac{\text{Total number of trees}}{\text{Total number of quadrats}} \)= \(\frac{25}{25}\) = 1

Standard Deviation \(\sigma(A) \):

\(\sigma(A) = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu(A))^2}\)

Explanation:

Forest Patch A:

There are 25 quadrats, each containing exactly 1 tree.

Mean (μA): Total number of trees is 25. Mean is \(\mu(A) = \frac{25}{25} = 1 \).

Standard deviation (σA): Since all values are the same (1), the standard deviation is \(\sigma(A) = 0\) .

Forest Patch B:

Quadrats contain the following numbers of trees: \(\{2, 0, 0, 0, 0, 0, 0, 7, 0, 10, 0, 0, 0, 6, 0\} \)

Mean (μB): Total number of trees is 2 + 7 + 10 + 6 = 25 . Mean is \(\mu(B) = \frac{25}{15} = 1.67 \) .

Standard deviation (σB): Since values vary greatly, the standard deviation \(\sigma(B) > 0\) .

⇒ \( \mu(A)\)  and \( \mu(B)\)  are not equal.

⇒ The standard deviation in Patch B is higher due to the large variation in tree numbers.

Thus, Option 3) is correct.

• The given series is: 150, 170, 155, 160, 180, 165, ?

• Start at 150
• Add 20 to get 170
• Subtract 15 to get 155
• Add 5 to get 160
• Add 20 to get 180
• Subtract 15 to get 165

• +20 (150 + 20 = 170)
• -15 (170 - 15 = 155)
• +5 (155 + 5 = 160)
• +20 (160 + 20 = 180)
• -15 (180 - 15 = 165)

Ascending order 1, 4, 9, 9, X, 14, 15, 15

Formula:

Median = Sum of two mid observation/2  [In case of even number of observaions]

Calculation

11 = (9 + x)/2

⇒ 22 – 9 = x

⇒ x = 13

Given:

Formula used:

Mean = \(\Sigma fixi \over \Sigma fi\)

where fi = frequency of particular value

xi = Mid value of the frequency class

Calculation: 

Mean = \(1670 \over 50\) = 33.4 

 ∴ The mean of the given frequency distribution table is 33.4.

GIVEN:

Five numbers 42, 24, 32, 64, 68

CONCEPT:

Concept of variance

FORMULA USED:

Mean = Sum/Total

Variance (σ²) = ∑δ²/n

CALCULATION:

Mean = (42 + 24 + 32 + 64 + 68)/5 = 230/5 = 46

∑δ² = |42 - 46|² + |24 - 46|² + |32 - 46|² + |64 - 46|² + |68 - 46|²

⇒ 16 + 484 + 196 + 324 + 484

⇒ 1504

Variance (σ²) = ∑δ²/n

⇒ 1504/5

= 300.8

Given:

Given observation: 3, 8, 4, 5, 9, 13

Concept used:

Standard deviation = \(\sigma = \sqrt {\frac{\sum x^2_i}{n}-{(\frac{\sum x_i}{n})}^2} \)

Calculation:

n = 6

\(\sum \frac{x_i}{n} = \frac {3 + 8 + 4 + 5 + 9 + 13}{6} \) = 7

\(\frac{\sum x^2_i}{n} = \frac {3^2 + 8^2 + 4^2 + 5^2 + 9^2 + 13^2}{6} \) = 60.66

Now, the standard deviation

\(\sqrt {{60.66} -7^2 } \)

\(3.41\)

The correct answer is 100.

Key Points

• Variance is the square of standard deviation.
• Here given the standard deviation of a population is 10.
• So, the population variance = 10² = 100.

Given:

Data: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Concept:

Mean: It is the average of given observation. Let x1, x2, …, xn be n observations, then

Mean  = \({\rm{\bar X}} \) = \(\dfrac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Mean deviation: Let x1, x2, …, xn be n observations, then:

Mean deviation  = \(\dfrac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right|}}{{\rm{n}}}\)

Calculation:

\(\sum \overline{x} \) = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 210

Mean \({\rm{\bar X}} = \frac{{210}}{20}\) = 10.5

⇒ X̅ = 10.5

Mean deviation  = \( \dfrac{{\mathop \sum \nolimits_{\rm{i}}^{\rm{n}} \left| {{{\rm{X}}_{\rm{i}}} - {\rm{\bar X}}} \right|}}{{\rm{n}}}\) 

\({{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right|}}\) = 8.5 + 6.5 + 4.5 + 2.5 + 0.5 + 1.5 + 3.5 + 5.5 + 7.5 + 9.5 + 9.5 + 7.5 + 5.5 + 3.5 + 1.5 + 0.5 + 2.5 + 4.5 + 6.5 + 8.5 = 100

⇒ \(\dfrac{{100}}{20}\) = 5

Given:

The standard deviation of a, b and c is t

Concept used:

If the standard deviation of any set of numbers is A 

And we are doing the same value mathematical operation to each of the terms, then the standard deviation will remains the same.

Calculation:

Here standard deviation of a, b and c is t

And we are adding 6 to every term

According to the concept, we have

Standard deviation of a + 6, b + 6 and c + 6 is t.

∴ The required standard deviation is t.

Given:

Numbers: 70, 55, 52, 85, 68, 67, 79

Formula used:

Mean = Sum of all observations / Total number of all observations

There are 'n' observations.

If n is odd, then the median is {(n + 1)/2}th term.

If n is even, then the median is the average of  (n/2)th term and {(n/2) + 1}th term.

Calculation:

Mean = \(\dfrac{70 + 55 + 52 + 85+ 68 + 67+ 79}{7}\)

⇒ \(\dfrac{476}{7}\) = 68

Arrange all the observations in ascending order.

52, 55, 67, 68, 70, 79, 85

n = 7

So, Median = {(7 + 1)/2}^th term

⇒ 4^the term = 68

Median = 68

The difference between their mean and median = 68 - 68 = 0

∴ The difference between their mean and median is 0.

Given: 

Concept Used:

Arithmetic Mean = ∑(f_i x_i )/n

Calculations:

Arithmetic mean = \(\frac{0\times 3+1\times 2+ 2 \times 2+3\times 4+ 4 \times 6+ 5\times 7+ 6\times 7+ 7\times5+8\times 3+ 9\times 4+ 10\times 2}{3+2+2+4+6+7+7+5+3+4+2}\)

Arithmetic mean = \(\frac{234}{45}=5.2\)

∴ The arithmetic mean of the given data is 5.2

Formula used:

The formula to calculate the mean deviation for the given data set is given below.

Mean Deviation = [Σ |X – µ|]/N

Here, 

Σ represents the addition of values

X represents each value in the data set

µ represents the mean of the data set

N represents the number of data values

| | represents the absolute value, which ignores the “-” symbol

Calculation:

Mean = (58 + 47 + 69 + 32 + 14)/5

⇒ 220/5 = 44

Deviation = (14 + 3 + 25 + 12 + 30)/5 

⇒ 16.8

∴ The correct answer is 16.8.