Question

Suppose that 100 items are drawn from a population f manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a right skewed probability distibution with mean=150 oz and standard deviation=30 oz. Find the value of X-bar for which P(X-bar > ?) = 0.75.

Answer 1

Solution:

Given that X follows a distribution with

= 150 and = 30

Given that this is a Right skewed distribution.

A random sample of size n = 100 is taken from this population.

Let be the mean of sample.

Since n = 100 > 30 , t he sampling distribution of the is approximately normal with

Mean = = 150

SD =      = 30/​100 = 3

Find the value of X-bar for which P(X-bar > ?) = 0.75

Suppose x be the required value .

So ,

P(X-bar > x) = 0.75

So , P((X-bar < x) = 1 - 0.75

P((X-bar < x) = 0.25

Form z table , see the probability 0.75 and then see corresponding z value .

So , P(Z < -0.674) = 0.25

So, z = -0.674

Now , using z score formula ,

x = + (z * )

= 150 + (-0.674 * 3)

= 147.98

The required value of  X-bar is 147.98