STANDARD DEVIATION WORKSHEET

1. Find the standard deviation and the coefficient of variation for the following numbers: 5, 8, 9, 2, 6.

2. Find the standard deviation for the following distribution :

3. Find the standard deviation of first n natural numbers.

4. Find the standard deviation of first 25 natural numbers.

5. If arithmetic mean and coefficient of variation of x are 10 and 40 respectively, what is the variance of (15 – 2x) ?

6. Calculate the mean and standard deviation of each data set

(a) 20, 21, 19, 22, 21, 20, 19, 20, 21, 20

(b) 303, 299, 306, 298, 304, 307, 299, 302, 305, 299, 300

7. The standard deviation of the weights (in kg) of the studensts of a class of 50 students was calculated to be 4.5 kg. Later on it was found that due to some fault in weighing machine, the weight of each student was under by 0.5 kg. The correct standard deviation of the weight will be

a)  less than 4.5    b) Greater than 4.5

c) Equal to 4.5      d) cannot be determined

8. If all observations in a distribution are increased by 6, then the variance of the series will be _____

a)  increased     b) decreased

c) unchanged      d) none of these

1. Answer :

Formula for standard deviation for an ungrouped data :

s = √(∑x_i²/n) - (∑x_i/n)²

s = √(210/5) - (30/5)²

s = √42 - 6²

s = √(42 - 36)

s = √6

s = 2.45

The coefficient of variation

= (Standard deviation/Arithmetic mean) x 100

= (2.45/6) x 100

= 40.83

The coefficient of variation is

2. Answer :

Formula for standard deviation for a grouped frequency distribution :

s = √(∑x_i²/n) - (∑x_i/n)²

s = √(2172/100) - (412/100)²

s = √21.72 - (4.12)²

s = √21.72 - 16.97

s = √4.7456

s = 2.18

So, the required standard deviation is 2.18.

3. Answer :

Let x_i = 1, 2, 3, ..........n.

Arithmetic mean of first n natural numbers :

Standard deviation of first n natural numbers :

4. Answer :

Standard deviation of first n natural numbers :

Substitute n = 25.

5. Answer :

Let y = 15 - 2x.

When x and y are related as y = a + bx, then

S_y = |b|S_x

S_y = |-2|S_x

S_y = 2S_x ----(1)

Coefficient of variation of x = 40 and mean of x = 10.

Coefficient of variation of x = 40

100 ⋅ (S_x/AM) = 40

100 ⋅ (S_x/10) = 40

10 ⋅ S_x = 40

S_x = 4

S_y = 2 ⋅ 4

S_y = 8

Variance of y = 8²

Variance of (15 - 2x) = 64

6. Answer :

Mean :

Mean = (20 + 21 + 19 + 22 + 21 + 20 + 19 + 20 + 21 + 20)/10

= 203/10

= 20.3

So, the required mean is 20.3

s = √(∑d²/n) - (∑d/n)²

s = √(10/10) - (3/10)²

= √1 - (0.3)²

= √1 - 0.09

= √0.91

= 0.95

So, the required standard deviation is 0.95.

(b) 303, 299, 306, 298, 304, 307, 299, 302, 305, 299, 300

Mean = (303+299+306+298+304+307+299+302+305+299+ 300)/11

= 3322/11

= 302

So, the required mean is 20.3

s = √(∑d²/n) - (∑d/n)²

s = √(102/11) - (0/11)²

= √(102/11)

= √9.27

= 3.04

So, the required standard deviation is 3.04

7. Answer :

Standard deviation remains unaffected due to a change of origin. So, correct standard deviation of 50 students is 4.5.

8. Answer :

Since, standard deviation is independent of teh change f origin therefore VARIANCE is also independent to change of origin. So, option c is correct.

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