Chapter 4.2, Problem 99E

(a)

To determine

To find out what percent of seeds weigh more than $500$ mg.

Answer to Problem 99E

  $59.10\%$ of seeds weigh more than $500$ mg.

Explanation of Solution

It is give in the question that,

  $\begin{array}{l}\mu =525\\ \sigma =110\\ x=500\end{array}$

The z -score is then calculated as:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{500-525}{110}\\ =\frac{-25}{110}\\ =-0.23\end{array}$

Thus the probability can be calculated as:

  $\begin{array}{l}P(X<500)=P(Z<-0.23)\\ =0.4090\end{array}$

Thus, now by using the complement rule we have,

  $\begin{array}{l}P(X>500)=1-P(X<500)\\ =1-0.4090\\ =0.5910\\ =59.10\%\end{array}$

(b)

To determine

To find out what is the smallest weight among the remaining seeds if we discard the lightest $10\%$ of these seeds.

Answer to Problem 99E

  $384.2$ milligrams.

Explanation of Solution

It is given in the question that,

  $\begin{array}{l}\mu =525\\ \sigma =110\end{array}$

Thus, first find out the z -score in table A that corresponds with a probability of $10\%$ as:

  $z=-1.28$

Then the z -score will be as:

  $z=\frac{x-\mu }{\sigma }$

Thus, by evaluating it we get,

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \Rightarrow z\sigma =x-\mu \\ \Rightarrow \mu +z\sigma =x\end{array}$

Thus, by solving this equation for finding the unknown value:

  $\begin{array}{l}x=\mu +z\sigma \\ =525-1.28(110)\\ =384.2\end{array}$

Then we know that the smallest weight among the remaining seeds if we discard the lightest $10\%$ of these seeds is $384.2$ milligrams.