Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 14.2 inches, and standard deviation of 4.4 inches.

If 45 items are chosen at random, what is the probability that their mean length is greater than 12.7 inches?

Answer 1

Solution:

Given that , X follows a distribution having

= 14.2

=  4.4

This distribution is skewed right.

A sample of size n = 45 is taken from this population.

Let be the mean of sample.

Since sample size n > 30 ,  sampling distribution of the is approximately normal.

= = 14.2

=      = 4.4/​45 =  0.6559132734

Now ,

P[sample mean is greater than 12.7 inches]

= P[ > 12.7]

= P[( - )/ >  (12.7 - )/]

= P[ ( - )/ > (12.7 - 14.2)/0.6559132734]

= P[Z > -2.29]

= 1 - P[Z < -2.29]

= 1 - 0.0110 ( use z table)

= 0.9890

Required Probability = 0.9890