Question

Find the probability that the Normal random variable with mean 21 and standard deviation 3.0 will generate an outlier (outside the inner fences) observation. Remember that the lower (upper) inner fence is 1.5*IQR below (above) the first (third) quartile.

a.

0.0035         

b.

0.0051         

c.

0.0058         

d.

0.0062         

e.

0.0070

Answer 1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 21

S.d = 3

First we need to find Q1 and Q3

We know that below Q3, there is 75%

From z table, P(z<0.67) = 75%

0.67 = (Q3 - 21)/3

Q3 = 23.01

We also know that below Q1, 25% lies

From z table, P(z<-0.67) = 25%

-0.67 = (Q1 - 21)/3

Q1 = 18.99

IQR = Q3-Q1 = 4.02

1.5*IQR = 6.03

Now outlier is above Q3 + (1.5*IQR) and Q1 - (1.5*IQR)

That is below 12.96 or greater than 29.04

P(12.96 < x < 29.04 ) = p(z<29.04) - p(z<12.96)

P(z<29.04) = 2.68

From z table, P(z<2.68) = 0.9963

P(z<12.96)

Z = (12.96 - 21)/3 = -2.68

From z table, P(z<-2.68) = 0.0037

= 0.9963 - 0.0037 = 0.9926

We want an outlier that is it will exist outside the interval

So.probability is 1 - 0.9926 = 0.0074 ~ 0.007