Chapter 6.1, Problem 29E

(a)

To determine

To Calculate: the event in terms of X and find its probability

Answer to Problem 29E

P (325<X<345) =0.8996

Explanation of Solution

Given:

  $\begin{array}{l}\mu =336\\ \sigma =6\\ x=325or345\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The event between 325 and 345 days can be represented in terms X as

325<X<345

Z score for X=325

  $z=\frac{325-336}{6}=-1.83$

Z score for X=345

  $z=\frac{345-336}{6}=1.50$

Corresponding probability using the normal probability table

P (325<X<345) =P (-1.83<Z<1.50)

  =P (Z<1.50)-P (Z<-1.83)

  =0.9332-0.0336

  =0.8996

  =89.36%

  

(b)

To determine

To Calculate: the value of c

Answer to Problem 29E

  $c=341.04$

Explanation of Solution

Given:

  $\begin{array}{l}\mu =336\\ \sigma =6\\ \text{P}\left(X\ge c\right)=0.20\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  $\begin{array}{c}\text{P}\left(X<c\right)=1-\text{P}\left(X\ge c\right)\\ =1-0.20\\ =0.80\end{array}$

  

By using graph and z table get the value of z

Z=0.84

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ 0.84=\frac{c-336}{6}\\ 5.04=c-336\\ c=336+5.04\\ c=341.04\end{array}$