Question

Cereal A company's cereal boxes advertise that each box contains $9.65$ ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean $\mu =9.70$ ounces and standard deviation $\sigma =0.03$ ounce.

a. What is the probability that a randomly selected box of the cereal contains less than $9.65$ ounces of cereal?

b. Now take an SRS of 5 boxes. What is the probability that the mean amount of cereal in these boxes is less than $9.65$ ounces?

Short answer

1. The resultant probability is $0.0475.$
   
2. The resultant probability is $0.0001$.

Step-by-step solution

Part (a) Step 1: Given information

Given:

The value of Population mean $\left(\mu \right)=9.70$

The value of Population standard deviation$\left(s\right)=0.03$

Part (a) Step 2: Calculation

Consider the random variable $X$, which represents the amount of cereal in a box.

The likelihood that a randomly chosen box contains less than $9.65$ ounces of cereal can be computed as follows:

$\begin{array}{rcl}P(X& <& 9.65)=P\left(\frac{x-\mu }{\sigma }<\frac{9.65-\mu }{\sigma }\right)\\ & =& P\left(Z<\frac{9.65-9.70}{0.03}\right)\\ & =& P(Z<-1.67)\\ & =& 0.0475\end{array}$

Thus, the required probability is $0.0475.$

Part (b) Step 1: Given information

The Sample size $\left(n\right)=5$

Part (b) Step 2: Calculation

The likelihood that the average amount of cereal in five randomly chosen boxes is less than $9.65$ounces can be computed as follows:

$\begin{array}{rcl}P(\overline{X}& <& 9.65)=P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{9.65-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ & =& P\left(Z<\frac{9.65-9.70}{\frac{0.03}{\sqrt{\overline{5}}}}\right)\\ & =& P(Z<-3.73)\\ & =& 0.0001\end{array}$