Question

A company’s single-serving cereal boxes advertise $1.63$ounces of cereal. In fact, the amount of cereal X in a randomly selected box can be modeled by a Normal distribution with a mean of $1.70$ounces and a standard deviation of $0.03$ounce. Let $Y=$the excess amount of cereal beyond what’s advertised in a randomly selected box, measured in grams ($1\mathrm{ounce}=28.35\mathrm{grams}$).

a. Find the mean of $Y$.

b. Calculate and interpret the standard deviation of $Y.$

c. Find the probability of getting at least $1gram$more cereal than advertised.

Short answer

a. The mean is $\mathbf{1}\mathbf{.}\mathbf{9845}$grams.

b. The excess amount of cereal differs by $\mathbf{0}\mathbf{.}\mathbf{8505}$grams on average from the mean excess amount of $1.9845$grams.

c. There's a $\mathbf{0}\mathbf{.}\mathbf{8770}\mathbf{}$chance of getting at least $1$gram more cereal than claimed.

Step-by-step solution

Part(a) Step 1 : Given Information 

Given :

Company’s Cereal box contains : $1.63$ounces of cereal.

Randomly selected box can be modeled by a Normal distribution with a mean of : $1.70$ounces

Standard deviation : $0.03$ounce

$Y=$the excess amount of cereal beyond what’s advertised

Part(a) Step 2 : Simplification  

Every data value in the $X$distribution is first subtracted by $1.63$and then multiplied by $28.35$.

The center of the distribution is also subtracted by the same constant if every data value is subtracted by the same constant.

Furthermore, if every data value is multiplied by the same constant, the distribution's center is multiplied by the same constant.

The mean is the center's measurement, as we all know.

As a result, the mean is reduced by $1.63$before being multiplied by $28.35$.

$$\begin{gathered} {\mu }_Y=28.35({\mu }_X–1.63) \\ =28.35(1.70–1.63) \\ =28.35(0.07) \\ =\mathbf{1}\mathbf{.}\mathbf{9845}\mathbf{}\mathit{g}\mathit{r}\mathit{a}\mathit{m}\mathit{s} \end{gathered}$$

Part(b) Step 1 : Given Information 

Given :

Company’s Cereal box contains : $1.63$ounces of cereal.

Randomly selected box can be modeled by a Normal distribution with a mean of :$1.70$ ounces

Standard deviation :$0.03$ounce

$Y=$the excess amount of cereal beyond what’s advertised

Part(b) Step 2 : Simplification  

The spread of the distribution is unaltered if every data value is removed by the same constant.

Furthermore, if every data value is multiplied by the same constant, the distribution's spread is also multiplied by the same constant.

The standard deviation is a measure of the spread, as we all know.

As a result, multiply the standard deviation by

$$\begin{gathered} {\sigma }_Y=28.350{\sigma }_X \\ =28.35(0.03) \\ =\mathbf{0}\mathbf{.}\mathbf{8505}\mathbf{}\mathit{g}\mathit{r}\mathit{a}\mathit{m}\mathit{s} \end{gathered}$$

Part(c) Step 1 : Given Information 

Given :

Company’s Cereal box contains : $1.63$ounces of cereal.

Randomly selected box can be modeled by a Normal distribution with a mean of : $1.70$ounces

Standard deviation : $0.03$ounce

$Y=$the excess amount of cereal beyond what’s advertised

Part(c) Step 2 : Simplification  

Calculate the $z-$score using the formula :

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{1-1.9845}{0.8505} \\ \approx \mathbf{-}\mathbf{1}\mathbf{.}\mathbf{16} \end{gathered}$$

To find the equivalent probability, use the normal probability table in the appendix. In the typical normal probability table for$P(z<-1.16)$, look for the row that starts with $-1.1$and the column that starts with $.06$.

$$\begin{gathered} P(x\ge 1)=P(z>-1.16) \\ =1-P(z<-1.16) \\ =1-0.1230 \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{8770} \end{gathered}$$