Standard Deviation Questions

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Standard Deviation questions and answers can help students learn the concepts fast. The standard deviation of a dataset is a measure of its dispersion related to its mean. Students can use these questions to get a thorough overview of the topics and practise solving them to deepen their understanding. To double-check your answers, read the entire explanations for each question. To learn more about standard deviation, click here.

Standard Deviation Definition

The measure of the dispersion of data points relative to the mean is defined by the standard deviation in descriptive statistics. It is a way of measuring the data points’ deviation from the mean and indicates how values are distributed across the data sample. The standard deviation of a set of data is equal to the square root of the variance.

Standard Deviation Questions with Solutions

Steps to Find Standard Deviation

Step 1: Determine the mean of the observations, i.e. Arithmetic mean.

Step 2: Calculate the squared deviations from the mean, i.e. (Data value – Mean)2

Step 3: Calculate the squared differences’ average, i.e. Variance = Sum of squared differences ÷ Total number of observations.

Step 4: To find the standard deviation, calculate the variance’s square root, i.e. Standard deviation = √Variance.

Check out Variance and Standard Deviation as well.

1. What is the best measure of dispersion, and how?

Answer:

The best measure of dispersion is standard deviation, which is why it is the most often used measure of dispersion:

  • It is dependent on all values, and so contains data for the entire series. As a result, even a small change in one variable has an impact on the standard deviation.
  • It is scale-independent but not origin-independent.
  • It’s useful for complex statistical calculations like comparing two data sets’ variability.
  • It can be applied to hypothesis testing.

2. The mean on a particular exam in a class of 100 was 50, and the standard deviation was 0. What does this imply?

Answer:

According to the definition of standard deviation, when the standard deviation of a series is 0, it means that all of the values in the series are equal to the mean, making all deviations zero, and hence, the standard deviation is also zero.

3. A garden contains 39 plants. The following plants were chosen at random, and their heights were recorded in cm: 38, 51, 46, 79, and 57. Calculate their heights’ standard deviation.

Solution:

Given that, Number of observations = 5

Hence, the mean of 5 observations is:

Mean = (38 + 51 + 46 + 79 + 57)/5 = 54.2

Now, the standard deviation is calculated as follows:

Standard Deviation, SD = √[(Σ(xi – x̄)2) / (N-1)]

Now, substitute the values in the formula, we get

\(\begin{array}{l}S.D = \sqrt{\frac{(51-54.2)^{2} + (38 – 54.2)^{2}+(79-54.2)^{2}+(46-54.2)^{2}+ (57-54.2)^{2}}{4}}\end{array} \)

On solving the above expression, we get

S.D = 15.5

Hence, the standard deviation (S.D) of their heights is 15.5.

\(\begin{array}{l}\text{4. Prove that the first n natural numbers’ standard deviation equals  }\sigma = \sqrt{\frac{n^{2}-1}{12}}.\end{array} \)

Solution:

The mean of first n natural numbers is calculated as follows.

Mean = Sum of all observations / Number of observations

Hence, mean is:

\(\begin{array}{l}= \frac{[n(n+1)]/2}{n}\end{array} \)

Hence, mean = (n+1)/2

As a result, the sum of the square of the first n natural number is:

\(\begin{array}{l}\sum x^{2} = \frac{n(n+1)(2n+1)}{6}\end{array} \)

Therefore, the standard deviation is:

\(\begin{array}{l}\sigma = \sqrt{\frac{\sum x^{2}}{n}-\left ( \frac{\sum x}{n} \right )^{2}} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\frac{n(n+1)(2n+1)}{6n}-\left ( \frac{n+1}{2} \right )^{2}} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\frac{(n+1)(2n+1)}{6}-\left ( \frac{n+1}{2} \right )^{2}} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\left ( \frac{n+1}{2} \right )\left [ \frac{2n+1}{3} – \frac{n+1}{2} \right ]} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\left ( \frac{n+1}{2} \right )\left [ \frac{2(2n+1)-3(n+1)}{6} \right ]} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\left ( \frac{n+1}{2} \right )\left [ \frac{4n+2-3n-3}{6} \right ]} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\left ( \frac{n+1}{2} \right )\left [ \frac{n-1}{6} \right ]} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\frac{n^{2}-1}{12}} \end{array} \)

Thus, first n natural numbers’ standard deviation,

\(\begin{array}{l}\sigma = \sqrt{\frac{n^{2}-1}{12}} \end{array} \)

Hence, proved.

5. Determine the variance and standard deviation of the random variable X with the following probability distribution:

X 0 1 2 3
P(X=x) 1/8 3/8 3/8 1/8

Solution:

Assume that the mean of the random variable X is μ.

\(\begin{array}{l}\mu = \sum_{x=0}^{3}X\times P(X)=12/8 = 3/2 \end{array} \)

Hence, the Var(X) is calculated as:

\(\begin{array}{l}\sigma ^{2} = \sum_{x=0}^{3}X^{2}\times P(X) – \mu ^{2} \end{array} \)
\(\begin{array}{l}\sigma ^{2} = \frac{24}{8}-\frac{9}{4}= \frac{3}{4} \end{array} \)

Now, the standard deviation is calculated as:

\(\begin{array}{l}\sigma = \sqrt{Var(x)} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\frac{3}{4}} \end{array} \)
\(\begin{array}{l}\sigma = \frac{\sqrt{3}}{2} \end{array} \)

6. Let X represent the total of the numbers obtained by rolling two fair dice. Calculate X’s variance and standard deviation.

Solution:

6 × 6 equals 36 observations are obtained, when two fair dice are rolled.

Hence, we get the following:

P(X=2)=P(1,1)= 1/36

​P(X=3)=P(1,2)+P(2,1)= 2/36 = 1/18

P(X=4)=P(1,3)+P(2,2)+P(3,1)= 3/36 = 1/12

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1) = 4/36 = 1/9

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1) = 5/36

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1) = 6/36 = 1/6

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2) = 5/36

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3) = 4/36 = 1/9

P(X=10)=P(4,6)+P(5,5)+P(6,4)= 3/36 = 1/12

P(X=11)=P(5,6)+P(6,5)= 2/36 = 1/18

P(X=12)=P(6,6) = 1/36

As a result, the required probability distribution looks like the following:

E(X)=∑Xi . P(Xi)

Now, substitute the values on the formula. We get,

= 2(1/36) + 3(1/18) + 4(1/12) + 5(1/9) + 6(5/36) + 7(⅙) + 8(5/36) + 9(1/9) + 10(1/12) + 11(1/18) + 12(1/36)

= (1/18) + (⅙) + (⅓) + (5/9) + (⅚) + (7/6) + (10/9) + 1 + (⅚) + (11/18) + (⅓)

Hence, E(X) =7

Now, find the values of E(X2) = ∑ Xi2.P(Xi)

Substituting the values in the formula, we get

E(X2) = 54.833.

Finding Variance:

Therefore, Var(X) = E(X2) – [E(x)]2

Now, substitute the obtained values in the formula. We get,

Var(X) = 54.833 – (7)2

Var(X) = 54.833 – 49

Var(X) = 5.833.

Finding Standard Deviation:

Therefore, the standard deviation is:

S.D = √5.833

S.D = 2.415.

7. P(X) is the probability density, with X being a discrete random variable. Find the standard deviation of x.

X 1 2 3
P(X) 0.3 0.6 0.1

Solution:

When it comes to discrete random variables, the mean can be found as follows:

Mean = E(X) = ∑XP(X)

Mean = 1(0.3) + 2(0.6) + 3(0.1)

Mean, E(X) = 1.8.

And E(X2) = ∑X2P(X) = 12(0.3) + 22(0.6) + 32(0.1)

E(X2) = 3.6.

Finding Variance:

Hence, Variance = E(X2) – [E(x)]2

Variance = 3.6 – (1.8)2

Variance = 0.36

Finding Standard Deviation:

Therefore, the standard deviation = √(0.36) = 0.6.

8. Using the actual mean method, calculate the standard deviation for the data 3, 2, 5, and 6.

Solution:

Given data observations are 3, 2, 5, and 6.

The number of observations = 4.

Finding Mean:

Now, the mean of the given data is:

Mean = (3+2+5+6)

Mean = 16/4

Mean = 4.

Finding Variance:

The squared differences from mean = (4-3)2+(2-4)2 +(5-4)2 +(6-4)2= 10

Variance = Squared differences from mean/Total number of data

Variance =10/4

Therefore, variance =2.5

Finding Standard Deviation:

S.D = √2.5

S.D = 1.58.

Therefore, the standard deviation is 1.58.

9. Determine the standard deviation of the first 5 natural numbers.

Solution:

As we already know, the standard deviation of first n natural numbers is

\(\begin{array}{l}\sigma = \sqrt{\frac{n^{2}-1}{12}}…(1) \end{array} \)

Now, we have to find the standard deviation of the first 5 natural numbers.

Therefore, substitute n = 5 in the equation (1)

\(\begin{array}{l}\sigma = \sqrt{\frac{5^{2}-1}{12}} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\frac{25-1}{12}} \end{array} \)
\(\begin{array}{l}\sigma = \sqrt{\frac{24}{12}} \end{array} \)
\(\begin{array}{l}S.D, \sigma = \sqrt{2} \end{array} \)

10. Find the standard deviation and variance for the following data: 10, 12, 8, 14, 16.

Solution:

Given data: 10, 12, 8, 14, 16.

Finding Mean:

Now, mean = (10 + 12 + 8 + 14 + 16)/5

Mean = 60/5

Mean = 12

Finding Variance:

Variance = [(10-12)2 + (12-12)2 + (8-12)2 +(14-12)2 + (16-12)2]/5

Variance = (4 + 0 + 16 + 4 + 16)/5

Variance = 40/5

Variance = 8

Finding Standard Deviation:

As we know, Standard deviation = √variance

Standard deviation = √8.

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Practice Questions

  1. A student who mistaken one observation for 50 instead of 40, computed the mean and standard deviation of 100 observations as 40 and 5.1, respectively. Find the exact standard deviation and mean.
  2. In a class of 50 pupils, four students were chosen at random, and their final evaluation scores were recorded as 812, 982, 836 and 769. Determine their standard deviation.
  3. With a standard deviation of Rs. 40, the average daily salary of 50 factory workers was Rs. 200. A rise of Rs. 20 has been offered to each employee. What is the new daily average salary and standard deviation?

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