Chapter 6.2, Problem 46E

a.

To determine

To find: The mean and standard deviation of Y

Answer to Problem 46E

The mean and standard deviation are 1.9845 and 0.8505.

Explanation of Solution

Given:

Population mean $\left(\mu \right)=9.70$

Population standard deviation $\left(\sigma \right)=0.03$

  $Z=28.35X$

Calculation:

Here, $Z=28.35X-273.0105$

The mean and standard deviation of Z are:

  $\begin{array}{c}{\mu }_Y=28.35\left(9.70\right)-273.0105\\ =1.9845\\ {\sigma }_Y=28.35\left(0.03\right)\\ =0.8505\end{array}$

Thus, the mean and standard deviation are 1.9845 and 0.8505 respectively.

b.

To determine

To find: The probability of getting at least more than 3-gram more cereal.

Answer to Problem 46E

The probability is 0.1170.

Explanation of Solution

Calculation:

The probability of getting at least more than 3-gram cereal can be calculated as:

  $\begin{array}{c}P\left(Z\ge 3\right)=P\left(Z\ge \frac{3-1.9845}{0.8505}\right)\\ =P\left(Z\ge 1.19\right)\\ =0.1170\end{array}$

Thus, the required probability is 0.1170.