A cereal manufacturer claims that the gross weight (including packaging) of a box of cereal labeled as weighing 12 oz has a mean of 12.2 oz and a standard deviation of 0.1 oz. You gather 75 boxes and weigh them all together. Let S denote the total weight of the 75 boxes of cereal.

a. If the claim is true, what is P(S ≤ 914.8)?

b. Based on the answer to part (a), if the claim is true, is 914.8 oz an unusually small total weight for a sample of 75 boxes?

c. If the total weight of the boxes were 914.8 oz, would you be convinced that the claim was false? Explain.

d. If the claim is true, what is P(S<910.3)?

e. Based on the answer to part (d), if the claim is true, is 910.3 oz an unusually small total weight for a sample of 75 boxes?

f. If the total weight of the boxes were 910.3 oz, would you be convinced that the claim was false? Explain.

Step 1 of 7</p>

Here we have to find the normal probabilities for the given conditions

The given mean

The given standard deviation is

Let S denote the weight of the 75 boxes

Then S is a normal variate with mean =75(12.2)

=915

Standard deviation=0.1

=0.866

Step 2 of 7</p>

a) Now we have to find

Then Z=

=(914.8-915)/0.866

=-0.231

Get the normal probabilities from the normal tables

=0.409

Step 3 of 7</p>

b) here =0.409

Means almost half of the samples are less than 914.8

So 914.8 is not a small total weight for a sample of 75 boxes

Step 4 of 7</p>

c) For testing the claim we need to do z- test

Here the test statistics is Z=

=

= -2

The critical value at 5% level of significance is -1.645

Here -2 the test statistics value falls in the critical region

So we have to reject the claim

Hence that the claim was false