Question

The Cheebles cookie factory changed their recipe. The inspectors took a sample of the new cookies and found that the sample was 42 grams with a standard deviation of 4 grams. the Cheebles CEO specially asked the inspectors to use these statistics to find the lower and upper boundary weighs of 50% of their cookies. What are the Z values of the limits of the limits of the area covering the middle half of the area under the normal curve that the inspectors would use to find this information for the CEO of Cheebles?

Answer 1

Solution:

Given:  The inspectors took a sample of the new cookies and found that the sample was 42 grams with a standard deviation of 4 grams.

That is Mean = and standard deviation = s = 4

We have  to use these statistics to find the lower and upper boundary weighs of 50% of their cookies.

That is find x values for middle 50% of the data.

First find the Z values of the limits of the area covering the middle half of the area under the normal curve.

That is find:

P( -z < Z < z ) = 50%

P( -z < Z < z ) = 0.50

If area between -z to +z is 0.50 then

area left of -z + area right of +z = 0.50

Thus area left of -z = 0.50 / 2 = 0.25

Now look in z table for Area = 0.2500 or its closest area and find z value.

Area 0.2514 is closest to 0.2500 and it corresponds to -0.6 and 0.07

thus z = -0.67

Since z ( standard normal distribution) is symmetric, P( Z < -0.67) = P( Z > 0.67)

that means z = 0.67

That is limits of z values of the area covering the middle half of the area under the normal curve are: ( -0.67 , 0.67).

Now use following formula to find x values:

and

Thus the lower and upper boundary weights of 50% of their cookies are:

Lower Boundary = 39.32 gram and

Upper Boundary = 44.68 grams.