Question

1. Suppose weights of the small version of Raisin Bran cereal boxes have a normal distribution with mean 10 ounces and standard deviation 3 ounces.
   1. A small cereal box is declared to be under filled if it weighs less than 2 standard deviations below the mean. What is the cutoff weight to be under filled?
   2. The top 10% of the boxes have a weight larger than what amount?
   3. What’s the chance that a randomly selected box of cereal weighs more than 11 ounces?

Answer 1

Solution:

Given: Weights of the small version of Raisin Bran cereal boxes have a normal distribution with mean 10 ounces and standard deviation 3 ounces.

X ~ Normal

Part a) small cereal box is declared to be under filled if it weighs less than 2 standard deviations below the mean.

That is z = -2.00

we have to find  the cutoff weight to be under filled.

Thus use following x value:

use following formula to find x value:

Thus the cutoff weight to be under filled is 4 ounces.

Part b) The top 10% of the boxes have a weight larger than what amount?

That is find x such that:

P( X > x ) = 10%

P( X > x ) = 0.1000

Thus find z such that:

P( Z > z ) = 0.1000

that is:

P( Z < z ) = 1 - P( Z > z )

P( Z < z ) = 1 - 0.1000

P( Z < z ) = 0.9000

Look in z table for area = 0.9000 or its closest area and find z value:

Area 0.8997 is closest to 0.9000 and it corresponds to 1.2 and 0.08

thus z = 1.28

Now use following formula to find x value:

ounces.

The top 10% of the boxes have a weight larger than 13.84 ounces.

Part c) What’s the chance that a randomly selected box of cereal weighs more than 11 ounces?

P( X> 11) = ...........?

Find z score for x = 11

Thus we get:

P( X > 11 ) = P( Z> 0.33)

P( X > 11 ) = 1 - P( Z< 0.33)

Look in z table for z = 0.3 and 0.03 and find corresponding area.

Thus

P(Z <0.33) = 0.6293

thus

P( X > 11 ) = 1 - P( Z< 0.33)

P( X > 11 ) = 1 - 0.6293

P( X > 11 ) = 0.3707