Question

1. the weights of seeds of a variety of beans are approximately normal with mean 535 mg and standard deviation 110 mg.

a. draw a picture of this normal distribution and allocate the 68%, 95% and 99.7% areas.
b. use the 68 - 95 - 99.7% rule to find the probability of choosing a seed between 315 mg and 645 mg.
c. use the 68 - 95 - 99.7% rule to find the weight of the largest 2.5% of seeds.
d. what percent of seeds weigh more than 500 mg?
e. in what weight range are the smallest 8% of seeds?
f. what is the probability of choosing a seed between 250 mg and 530mg?

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

Answer:

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