Question

9.Find the mean, variance and standard deviation using short-cut methodHeight 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115in cmsNo. of 3 4 7 7children15 96

Answer 1

The given data is:

Height in cms	70-75	75-80	80-85	85-90	90-95	95-100	100-105	105-110	110-115
Number of children	3	4	7	7	15	9	6	6	3

The midpoint of the intervals is calculated by adding first and last terms and dividing it by 2. The formula to calculate the new values of the given data is,

y i = x i −A h

Here, A is the assumed mean and h is the height between the classes.

Substitute 92.5 for A and 5 for h in the above formula and make a table by using these values.

Height in cms	Number of children, f i	Midpoint, x i	y i = x i −A h	( y i ) 2	f i y i	f i ( y i ) 2
70-75	3	72.5	−4	16	−12	48
75-80	4	77.5	−3	9	−12	36
80-85	7	82.5	−2	4	−14	28
85-90	7	87.5	−1	1	−7	7
90-95	15	92.5	0	0	0	0
95-100	9	97.5	1	1	9	9
100-105	6	102.5	2	4	12	24
105-110	6	107.5	3	9	18	54
110-115	3	112.5	4	16	12	48
	∑ i=1 n f i =60				∑ i=1 n y i f i =6	∑ i=1 n f i ( y i ) 2 =254

The formula to calculate the mean is,

x ¯ =A+ ∑ i=1 n y i f i ∑ i=1 n f i ×h x ¯ =A+ ∑ i=1 n y i f i N ×h (1)

Where, N is the sum of frequency.

Substitute 92.5 for A, 5 for h, 60 for N and 6 for ∑ i=1 n y i f i in equation (1).

x ¯ =92.5+ 6 60 ×5 =92.5+0.5 =93

Therefore, the mean of the given data is 93.

The formula to calculate the variance is,

σ 2 = h 2 N 2 [ N ∑ i=1 n f i ( y i ) 2 − ( ∑ i=1 n f i y i ) 2 ](2)

Substitute 60 for N, 5 for h, 6 for ∑ i=1 n y i f i and 254 for ∑ i=1 n f i ( y i ) 2 in equation (2).

σ 2 = ( 5 ) 2 ( 60 ) 2 [ 60( 254 )− ( 6 ) 2 ] = 25 3600 ( 15240−36 ) = 25 3600 ( 15204 ) =105.58

Thus, the variance of the given data is 105.58.

The formula to calculate the standard deviation is,

S.D.= σ 2 (3)

Substitute 105.58 for σ 2 in equation (3).

S.D.= 105.58 =10.27

Thus, standard deviation of the given data is 10.27.