1. the weights of seeds of a variety of beans are approximately normal with mean 535 mg and standard…

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu = 535$ is the mean and $\sigma=110$ is the standard deviation.

Step2: For part D

First, find the z - score for $x = 500$: $z=\frac{500 - 535}{110}=\frac{- 35}{110}\approx - 0.32$. Using the standard normal distribution table, the area to the left of $z=-0.32$ is approximately $0.3745$. The percentage of seeds weighing more than $500$ mg is $P(X>500)=1 - 0.3745 = 0.6255$ or $62.55%$.

Step3: For part E

We want to find the value of $x$ such that $P(X < x)=0.08$. Looking up the z - score in the standard - normal table corresponding to an area of $0.08$, we get $z\approx - 1.405$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we can solve for $x$: $-1.405=\frac{x - 535}{110}$. $x-535=-1.405\times110$. $x=535-1.405\times110=535 - 154.55 = 380.45$ mg. So the smallest $8%$ of seeds weigh less than $380.45$ mg.

Step4: For part F

Find the z - score for $x_1 = 250$: $z_1=\frac{250 - 535}{110}=\frac{-285}{110}\approx - 2.59$. Find the z - score for $x_2 = 530$: $z_2=\frac{530 - 535}{110}=\frac{-5}{110}\approx - 0.05$. Using the standard - normal table, $P(Z < - 2.59)\approx0.0048$ and $P(Z < - 0.05)\approx0.4801$. $P(-2.59<Z < - 0.05)=0.4801 - 0.0048 = 0.4753$ or $47.53%$.

Answer:

D. $62.55%$ E. Weights less than $380.45$ mg F. $47.53%$