Question

assume that the amounts of weight that male college students gain during the
with a mean of \\( \mu = 1.1 \mathrm { kg } \\) and a standard deviation of \\( \sigma = 4.6 \mathrm { kg } \\). complete parts

a. if 1 male college student is randomly selected, find the probability that he gains
freshman year.

the probability is 0.2539
(round to four decimal places as needed.)

b. if 9 male college students are randomly selected, find the probability that their mean
year is between 0 kg and 3 kg.

the probability is \\( \square \\)
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The standard error $\sigma_{\bar{x}}$ for a sample of size $n = 9$ is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Substituting $\sigma = 4.6$ kg and $n = 9$, we get $\sigma_{\bar{x}}=\frac{4.6}{\sqrt{9}}=\frac{4.6}{3}\approx1.5333$ kg.

Step2: Calculate the z - scores

For $x_1 = 0$ kg:
The z - score $z_1=\frac{x_1-\mu}{\sigma_{\bar{x}}}=\frac{0 - 1.1}{1.5333}\approx - 0.72$.
For $x_2 = 3$ kg:
The z - score $z_2=\frac{x_2-\mu}{\sigma_{\bar{x}}}=\frac{3 - 1.1}{1.5333}=\frac{1.9}{1.5333}\approx1.24$.

Step3: Find the probabilities

Using the standard normal distribution table, $P(Z\lt - 0.72)=0.2358$ and $P(Z\lt1.24) = 0.8925$.
The probability $P(0\lt\bar{X}\lt3)=P(Z\lt1.24)-P(Z\lt - 0.72)$.
Substituting the values, we get $P(0\lt\bar{X}\lt3)=0.8925 - 0.2358=0.6567$.

Answer:

$0.6567$