Question

10.The diameters of circles (in mm) drawn in a design are given belov49-52Diameters 33-36 37-40 41-44 45-48No. of circles 15172125

Answer 1

The given data is:

Diameters	33-36	37-40	41-44	45-48	49-52
Number of circles	15	17	21	22	25

Since the given data is not continuous, therefore it has to be made continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of ach class interval. 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 42.5 for A and 4 for h in the above formula and make a table by using these values.

Diameters	Number of circles, f i	Midpoint, x i	y i = x i −A h	( y i ) 2	f i y i	f i ( y i ) 2
32.5-36.5	15	34.5	−2	4	−30	60
36.5-40.5	17	38.5	−1	1	−17	17
40.5-44.5	21	42.5	0	0	0	0
44.5-48.5	22	46.5	1	1	22	22
48.5-52.5	25	50.5	2	4	50	100
	∑ i=1 n f i =100				∑ i=1 n y i f i =25	∑ i=1 n f i ( y i ) 2 =199

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 42.5 for A, 4 for h, 100 for N and 25 for ∑ i=1 n y i f i in equation (1).

x ¯ =42.5+ 25 100 ×4 =42.5+1 =43.5

Therefore, the mean of the given data is 43.5.

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 100 for N, 4 for h, 25 for ∑ i=1 n y i f i and 199 for ∑ i=1 n f i ( y i ) 2 in equation (2).

σ 2 = ( 4 ) 2 ( 100 ) 2 [ 100( 199 )− ( 25 ) 2 ] = 16 10000 ( 19900−625 ) = 16 10000 ( 19275 ) =30.84

Thus, the variance of the given data is 30.84.

The formula to calculate the standard deviation is,

S.D.= σ 2 (3)

Substitute 30.84 for σ 2 in equation (3).

S.D.= 30.84 =5.55

Thus, standard deviation of the given data is 5.55.