Chapter 4.2, Problem 91E

(a)

To determine

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

Answer to Problem 91E

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

Explanation of Solution

It is given that:

  $\begin{array}{l}\mu =\text{Mean=525}\\ \sigma =\text{Standard deviation=110}\\ x=500\end{array}$

The z -score is the value decreased by the mean divided by the standard deviation.

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

The probability to the left of the z -score is then given in the row with $-0.2$ and in the column with $0.03$ of the normal probability table in the appendix:

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

Now, using the complement rule. It is as follows,

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

Thus, $59.10\%$ of seeds weighs more than $500$ mg.

(b)

To determine

To find out what is the smallest weight among the remaining seeds.

Answer to Problem 91E

The smallest weight among the $90\%$ heaviest seeds is $384.20$ mg.

Explanation of Solution

It is given that:

  $\begin{array}{l}\mu =\text{Mean=525}\\ \sigma =\text{Standard deviation=110}\\ x=500\end{array}$

Now calculate z -score in table A that corresponds with a probability of $10\%\text{ or }0.10$ :

  $z=-1.28$

The z -score is the value decreased by the mean divided by the standard deviation:

  $z=\frac{x-\mu }{\sigma }$

Multiply each side by $\sigma$ :

  $z\sigma =x-\mu$

Add $\mu$ to each side:

  $\mu +z\sigma =x$

Thus we then have derived the formula $\mu +z\sigma =x$ . Filling in the known values then results in:

  $\begin{array}{l}x=\mu +z\sigma \\ =525-1.28(110)\\ =384.20\end{array}$

Then, we know that the smallest weight among the $90\%$ heaviest seeds is $384.20$ mg.