Question

assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of \\( \mu = 1.4 \mathrm{kg} \\) and a standard deviation of \\( \sigma = 4.8 \mathrm{kg} \\). complete parts (a) through (c) below.
a. if 1 male college student is randomly selected, find the probability that he gains between 0 kg and 3 kg during freshman year.
the probability is.2434.
(round to four decimal places as needed.)
b. if 9 male college students are randomly selected, find the probability that their mean weight gain during freshman year is between 0 kg and 3 kg.
the probability is \\( \square \\).
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard deviation for the sample mean

For a sample of size \(n = 9\), the standard deviation of the sample mean \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 4.8\) kg and \(n = 9\), then \(\sigma_{\bar{x}}=\frac{4.8}{\sqrt{9}}=\frac{4.8}{3}=1.6\) kg.

Step2: Standardize the values

We use the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\).
For \(\bar{x}_1 = 0\), \(z_1=\frac{0 - 1.4}{1.6}=\frac{- 1.4}{1.6}=-0.875\)
For \(\bar{x}_2 = 3\), \(z_2=\frac{3 - 1.4}{1.6}=\frac{1.6}{1.6}=1\)

Step3: Find the probability using the standard normal distribution

We want to find \(P(0<\bar{X}<3)\), which is equivalent to \(P(-0.875<Z<1)\)
Using the standard - normal table, \(P(Z < 1)=0.8413\) and \(P(Z<-0.875)=0.1908\)
Then \(P(-0.875 < Z < 1)=P(Z < 1)-P(Z<-0.875)\)
\(P(-0.875 < Z < 1)=0.8413-0.1908 = 0.6505\)

Answer:

\(0.6505\)