Question

the population of weights of a particular fruit is normally distributed, with a mean of 575 grams and a standard deviation of 8 grams. if 23 fruits are picked at random, then 18% of the time, their mean weight will be greater than how many grams? round your answer to the nearest gram.

Explanation:

Step1: Find the z - score

We know that if \(P(X>x)=0.18\), then \(P(X\leq x) = 1 - 0.18=0.82\).
Using the standard normal distribution table (or a calculator with a normal - distribution function, such as the invNorm function on a TI - 84), the \(z\) - score corresponding to a cumulative probability of \(0.82\) is \(z\approx0.92\).

Step2: Use the formula for the z - score of the sample mean

The formula for the \(z\) - score of the sample mean \(\bar{X}\) is \(z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
We are given \(\mu = 575\), \(\sigma = 8\), and \(n = 23\).
We want to solve for \(\bar{X}\) in the equation \(z=\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
Substitute the known values: \(0.92=\frac{\bar{X}-575}{\frac{8}{\sqrt{23}}}\).

First, calculate \(\frac{8}{\sqrt{23}}\approx\frac{8}{4.796}\approx1.668\).
Then, rewrite the equation as \(\bar{X}-575=0.92\times1.668\).
\(0.92\times1.668 = 1.53456\).
\(\bar{X}=575 + 1.53456\).

Answer:

\(577\) grams