Question

assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of $mu = 1.2$ kg and a standard - deviation of $sigma = 5.7$ 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 0.2087
(round to four decimal places as needed.)
b. if 16 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
(round to four decimal places as needed.)

Explanation:

Step1: Calculate z - scores for sample mean

The formula for the z - score of the sample mean is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\mu = 1.2$, $\sigma = 5.7$, and $n = 16$.
For $\bar{x}=0$:
$z_1=\frac{0 - 1.2}{\frac{5.7}{\sqrt{16}}}=\frac{- 1.2}{\frac{5.7}{4}}=\frac{-1.2}{1.425}\approx - 0.84$
For $\bar{x}=3$:
$z_2=\frac{3 - 1.2}{\frac{5.7}{\sqrt{16}}}=\frac{1.8}{\frac{5.7}{4}}=\frac{1.8}{1.425}\approx1.26$

Step2: Find probabilities using z - table

We want to find $P(-0.84$P(-0.84 < Z<1.26)=P(Z < 1.26)-P(Z < - 0.84)$
From the standard normal table, $P(Z < 1.26)=0.8962$ and $P(Z < - 0.84)=0.2005$.
So $P(-0.84 < Z<1.26)=0.8962-0.2005 = 0.6957$

Answer:

$0.6957$