Question

7. Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210Frequencies 23510

Answer 1

The given data is:

Classes	0-30	30-60	60-90	90-120	120-150	150-180	180-210
frequencies	2	3	5	10	3	5	2

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 105 for A and 30 for h in the above formula and make a table by using these values.

Classes	Frequencies, f i	Midpoint, x i	y i = x i −A h	( y i ) 2	f i y i	f i ( y i ) 2
0-30	2	15	−3	9	−6	18
30-60	3	45	−2	4	−6	12
60-90	5	75	−1	1	−5	5
90-120	10	105	0	0	0	0
120-150	3	135	1	1	3	3
150-180	5	165	2	4	10	20
180-210	2	195	3	9	6	18
	∑ i=1 n f i =30				∑ i=1 n y i f i =2	∑ i=1 n f i ( y i ) 2 =76

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 105 for A, 30 for h, 30 for N and 2 for ∑ i=1 n y i f i in equation (1).

x ¯ =105+ 2 30 ×30 =105+2 =107

Therefore, the mean of the given data is 107.

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 30 for N, 30 for h, 2 for ∑ i=1 n y i f i and 76 for ∑ i=1 n f i ( y i ) 2 in equation (2).

σ 2 = ( 30 ) 2 ( 30 ) 2 [ 30( 76 )− ( 2 ) 2 ] =1( 2280−4 ) =2276

Thus, the variance of the given data is 2276.