Question

The labels on boxes of Cheesy Poofs say that the box contains 16 ounces of Cheesy Poofs, but that doesn’t guarantee that each box contains exactly 16.00 ounces. The machine that fills the boxes doesn’t always put out the exact same amount. The amount in each box is Normally distributed, and has a standard deviation is .40 ounces. To help assure that most boxes contain as much or more than the label says, the machine is set so that the mean will be 16.28 ounces. What proportion of the boxes contain…                                           

1.    …less than 16.00 ounces?

2.    …between 16.00 and 17.00 ounces?

3. …more than 16.50 ounces?

Answer 1

Solution:

Given: The amount in each box is Normally distributed with the mean of 16.28 ounces and a standard deviation of 0.40 ounces.

That is:

Part a) Find the proportion of the boxes contain less than 16.00 ounces.

That is:

P( X < 16.00) =.........?

Find z score:

Thus we get:

P( X < 16.00) = P( Z < -0.70)

Look in z table for z = -0.7 and 0.00 and find area.

P( Z< -0.70) = 0.2420

Thus

P( X < 16.00) = P( Z < -0.70)

P( X < 16.00) = 0.2420

Thus the proportion of the boxes contain less than 16.00 ounces is 0.2420.

Part b) Find the proportion of the boxes contain between 16.00 and 17.00 ounces.

That is:

P( 16.00 < X < 17.00) =............?

Find z score for x = 16.00 and for x = 17.00

Thus we get:

P( 16.00 < X < 17.00) = P( -0.70 < Z < 1.80)

P( 16.00 < X < 17.00) = P( Z < 1.80) - P( Z< -0.70 )

We have : P( Z < -0.70 ) = 0.2420

Look in z table for z = 1.8 and 0.00 and find area.

P( Z< 1.80) = 0.9641

Thus we get:

P( 16.00 < X < 17.00) = P( Z < 1.80) - P( Z< -0.70 )

P( 16.00 < X < 17.00) = 0.9641 - 0.2420

P( 16.00 < X < 17.00) = 0.7221

Thus the proportion of the boxes contain between 16.00 and 17.00 ounces is 0.7221

Part c) Find the proportion of the boxes contain more than 16.50 ounces

P( X > 16.50 ) = ..............?

Find z score for x = 16.50

Thus we get:

P( X > 16.50 ) = P( Z > 0.55)

P( X > 16.50 ) = 1 - P( Z < 0.55)

Look in z table for z = 0.5 and 0.05 and find area.

P( Z< 0.55 ) = 0.7088

thus

P( X > 16.50 ) = 1 - P( Z < 0.55)

P( X > 16.50 ) = 1 - 0.7088

P( X > 16.50 ) = 0.2912

Thus the proportion of the boxes contain more than 16.50 ounces is 0.2912