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Chapter 6: Q. 46 (page 379)

46. Cereal A company's single-serving cereal boxes advertise 9.63 ounces of cereal. In fact, the amount of cereal X in a randomly selected box follows a Normal distribution with a mean of 9.70 ounces and a standard deviation of 0.03 ounces.
(a) LetY=the excess amount of cereal beyond what's advertised in a randomly selected box, measured in grams ( 1 ounce =28.35grams). Find the mean and standard deviation of Y.
(b) Find the probability of getting at least 3 grams more cereal than advertised.

Short Answer

Expert verified

(a) The mean and standard deviation of Yare 1.9845and 0.8505.

(b) The probability of getting at least 3grams more cereal than advertised is 0.1170.

Step by step solution

01

Part (a) Step 1: Given information

The excess amount of cereal beyond what's advertised in a randomly selected box, measured in grams is considered asY.

Where, 1ounce =28.35grams.

02

Part (a)  Step 2: Explanation 

According to the information the population mean (渭)=9.70
Population standard deviation(蟽)=0.03
The value of Z=28.35X
And then, Z=28.35X-273.0105
The mean and standard deviation ofZ are:
渭Y=28.35(9.70)鈭�273.0105=1.9845蟽Y=28.35(0.03)=0.8505

03

Part (b) Step 1: Given information 

The probability of getting at least 3grams more cereal than advertised.

04

Part (b) Step 2: Explanation 

The probability of getting at least more than3-gram cereal is:
P(Z鈮�3)=PZ鈮�3鈭�1.98450.8505=P(Z鈮�1.19)=0.1170

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