Seed weights $(2.2)$ Biological measurements on the same species often follow a Normal distribution quite closely. The weights of seeds of a variety of winged beans are approximately Normal with a mean of $525$ milligrams (mg) and a standard deviation of 110 mg.

(a) What percent of seeds weigh more than $500$ mg? Show your method.

(b) If we discard the lightest $10\%$ of these seeds, what is the smallest weight among the remaining seeds? Show your method.

The meanweights of seeds, $\overline{x}=525mg$

The standard deviation, $s=110mg$

Biological measurements of the same species are considered to frequently follow a Normal distribution.

A simple random sample (SRS) of size $n$ is made up of $n$ people chosen from the population with an equal chance of being the sample that is actually chosen.

It is given in the question that,

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

The $z-$score is calculated as follows:

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

Thus the probability can be calculated as:

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

As a result of using the complement rule, we now have

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

It is stated in the question, $$\begin{gathered} \mu =525 \\ \sigma =110 \end{gathered}$$

To begin, get the$z-$score in table A that corresponds to a $10\%$ likelihood as follows:

$z=-1.28$

Then the $z-$score will be as:

$z=\frac{x-\mu }{\sigma }$

Thus, by evaluating it we get,$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ \Rightarrow z\sigma =x-\mu \\ \Rightarrow \mu +z\sigma =x \end{gathered}$$

As a result of solving this equation for the unknown value: $$\begin{gathered} x=\mu +z\sigma \\ =525-1.28\left(110\right) \\ =384.2 \end{gathered}$$

After discarding the lightest $10\%$ of the seeds, we know that the smallest weight among the remaining seeds is $384.2$ milligrams.