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

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)

Concept:

The Standard Deviation: The Standard Deviation is the most stable measure of variability. Therefore it is most commonly used in research studies.

Calculation of Standard Deviation for Ungrouped Data

• Standard Deviation for ungrouped data can be computed by the following formula.
• The formula for Standard Deviation = \(\sqrt{\frac{{∑ X^2}}{N}}\)

Where x = Deviation from mean of the data, N = Total number of students.

Given:

Marks of students: 85, 65, 55, 42, 69, 83, 92, 77.

Calculation:

Mean of the marks = \(\frac{85~+~65~+~55~+~42~+~69~+~83~+~92~+~77}{8}\)

\(=\frac{568}{8}=71\)

We know that,

Standard Deviation (SD) = \(\sqrt{\frac{{∑ X^2}}{N}}\)

Here \(\sum x^2\) = 1954, N = 8

Then,

SD = \(\sqrt{\frac{{1954}}{8}}\)

= \(\sqrt{244.25}\) = 15.63

Thus the standard deviation for this data is 15.63

Additional Information

Steps For ungrouped data

1. Add all the scores (∑x) and divide this sum by the number of scores (N) and find out the mean.
2. Find out the difference between scores and means (X - x) and find out deviation (x).
3. Square all the deviation to get x².
4. Add all the squared deviation to get \(\sum x^2\)
5. Divide \(\sum x^2\) by N.
6. Find out the square root of the obtained values.

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:

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:

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: 

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

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.

The Standard Deviation: The Standard Deviation is the most stable measure of variability. Therefore it is most commonly used in research studies.

Calculation of Standard Deviation for Ungrouped Data

• Standard Deviation for ungrouped data can be computed by the following formula.
• Formula SD =\(\sqrt{\frac{({\sum X - \bar{X}})^2}{N}}\)

The above formula can be explained by the following example.

Mean= ∑X/N = 480/8 = 60

To calculate standard deviation using the formula, we get the following

SD= √(∑x2 ) /N

Here ∑x2 =382 N = 8 SD= √ 382/8 = √ 47.7 = 6.91

Thus the standard deviation for this data is 6.91.

Additional Information

Steps For ungrouped data

1. Add all the scores (∑x) and divide this sum by the number of scores (N) and find out the mean.
2. Find out the difference between scores and means (X-x) and find out deviation (x).
3. Square all the deviation to get x2.
4. Add all the squared deviation to get∑x2.
5. Divide ∑x2 by N.
6. Find out the square root of the obtained values.

Hence, we can conclude that the correct option is 3.