Question

assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of μ = 1.3 kg and a standard deviation of σ = 5.5 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 2165
(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 5849
(round to four decimal places as needed.)

c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

a. since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.
b. since the weight gain exceeds 30, the distribution of sample means is a normal distribution for any sample size.
c. since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size.
d. since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size.

Explanation:

Step1: Standardize for part a

For a normal - distribution $X\sim N(\mu = 1.3,\sigma = 5.5)$. We use the formula $Z=\frac{X-\mu}{\sigma}$.
For $X = 0$, $Z_1=\frac{0 - 1.3}{5.5}\approx - 0.24$.
For $X = 3$, $Z_2=\frac{3 - 1.3}{5.5}\approx0.31$.
$P(0Since $\Phi(-z)=1 - \Phi(z)$, $P(-0.24Using the standard - normal table, $\Phi(0.31)=0.6217$ and $\Phi(0.24)=0.5948$.
$P(-0.24

Step2: Standardize for part b

The sampling distribution of the sample mean $\bar{X}\sim N(\mu_{\bar{X}}=\mu,\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}})$ when the population is normally distributed. Here, $n = 9$, $\mu = 1.3$, $\sigma = 5.5$, so $\sigma_{\bar{X}}=\frac{5.5}{\sqrt{9}}=\frac{5.5}{3}\approx1.833$.
For $\bar{X}=0$, $Z_1=\frac{0 - 1.3}{1.833}\approx - 0.71$.
For $\bar{X}=3$, $Z_2=\frac{3 - 1.3}{1.833}\approx0.93$.
$P(0<\bar{X}<3)=P(-0.71Since $\Phi(-z)=1 - \Phi(z)$, $P(-0.71Using the standard - normal table, $\Phi(0.93)=0.8238$ and $\Phi(0.71)=0.7611$.
$P(-0.71

Step3: Answer part c

The Central Limit Theorem states that if the original population is normally distributed, the distribution of sample means is a normal distribution for any sample size.

Answer:

a. $0.2165$
b. $0.5849$
c. A. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.