Question

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) Let$Y=$the excess amount of cereal beyond what's advertised in a randomly selected box, measured in grams ( $1$ ounce $=28.35$grams). Find the mean and standard deviation of $Y$.
(b) Find the probability of getting at least $3$ grams more cereal than advertised.

Short answer

(a) The mean and standard deviation of $Y$are $1.9845$and $0.8505$.

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

Step-by-step solution

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 as$Y$.

Where, $1$ounce $=28.35$grams.

Part (a)  Step 2: Explanation 

According to the information the population mean $\left(\mu \right)=9.70$
Population standard deviation$\left(\sigma \right)=0.03$
The value of $Z=28.35X$
And then, $Z=28.35X-273.0105$
The mean and standard deviation of$Z$ are:
$$\begin{gathered} \begin{array}{c}{\mu }_Y=28.35\left(9.70\right)-273.0105\end{array} \\ =1.9845 \\ {\sigma }_Y=28.35\left(0.03\right) \\ =0.8505 \end{gathered}$$

Part (b) Step 1: Given information 

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

Part (b) Step 2: Explanation 

The probability of getting at least more than$3$-gram cereal is:
$$\begin{gathered} \begin{array}{c}P(Z\ge 3)=P\left(Z\ge \frac{3-1.9845}{0.8505}\right)\end{array} \\ =P(Z\ge 1.19) \\ =0.1170 \end{gathered}$$