Question

The observations are listed : 5.32, 9.87, 11.25, 10.94, 5.58, 6.29, 7.47, 10.75, 6.22, 8.00. Apply backward empirical rule, IQR/S, and normal probability plot to check the normality assumption.

Answer 1

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean.

mean = 8.169

Standard deviation =2.101

3rd rule, =

= (1.866 , 14.472)

All the data are in 3 standard deviation limit.

2nd rule,

= (3.967 , 12.371)

All the data are in 2 standard deviation limit. As we have only 10 observation.

1st rule,

= (6.07 , 10.27)

50% of the data are in 1 standard deviation limit.

We arrange the dataset in ascending order.

5.32 , 5.58, 6.22, 6.29, 7.47, 8.00, 9.87, 10.75, 10.94 , 11.25

Q1 (first quartile)= N /4 th term = 10/4 th term = 2.5 th term = 3rd term = 6.22

Q3 (3rd Quartile)= 3N /4 th term = 30/4 th term = 7.5 th term = 8th term = 10.75

IQR = Q3-Q1 = 10.75 - 6.22 = 4.53

S= standard deviation

=[ [(5.32-8.169)2 + (5.88-8.169)2 +...+ (11.25-8.169)2] /10 ]1/2= 2.101

IQR / S = 4.53/2.101 = 2.156

Normal Probability plot

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