Question

assume that the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of μ = 1.4 kg and a standard deviation of σ = 4.8 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 .6505. (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?
o a. since the distribution is of sample means, not individuals, the distribution is a normal distribution for any sample size.
o b. since the distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size.
o c. since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.
o d. since the weight gain exceeds 30, the distribution of sample means is a normal distribution for any sample size.

Explanation:

Step1: Calculate the z - scores for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 0\), \(z_1=\frac{0 - 1.4}{4.8}=\frac{-1.4}{4.8}\approx - 0.29\).
For \(x = 3\), \(z_2=\frac{3 - 1.4}{4.8}=\frac{1.6}{4.8}\approx0.33\).

Step2: Find the probabilities using the standard normal table

\(P(0<X<3)=P(-0.29<Z<0.33)\).
We know that \(P(Z < 0.33)=0.6293\) and \(P(Z<-0.29) = 0.3859\).
So \(P(-0.29<Z<0.33)=P(Z < 0.33)-P(Z<-0.29)\).
\(P(-0.29<Z<0.33)=0.6293 - 0.3859=0.2434\).

Step3: For part (b), calculate the standard error

The standard error \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(n = 9\) and \(\sigma=4.8\). So \(\sigma_{\bar{x}}=\frac{4.8}{\sqrt{9}}=\frac{4.8}{3}=1.6\).

Step4: Calculate the z - scores for the sample mean

For \(\bar{x}=0\), \(z_1=\frac{0 - 1.4}{1.6}=\frac{-1.4}{1.6}=-0.875\).
For \(\bar{x}=3\), \(z_2=\frac{3 - 1.4}{1.6}=\frac{1.6}{1.6}=1\).

Step5: Find the probabilities using the standard normal table

\(P(0<\bar{X}<3)=P(-0.875<Z<1)\).
\(P(Z < 1)=0.8413\), \(P(Z<-0.875)=0.1908\).
\(P(-0.875<Z<1)=P(Z < 1)-P(Z<-0.875)=0.8413-0.1908 = 0.6505\).

Step6: For part (c)

Since the original population (weight gains of male college students) is normally distributed (\(\mu = 1.4\) and \(\sigma=4.8\)), the sampling distribution of the sample mean \(\bar{X}\) is also normally distributed for any sample size \(n\) (by the property of normal distributions: if \(X\sim N(\mu,\sigma^{2})\), then \(\bar{X}\sim N(\mu,\frac{\sigma^{2}}{n})\)).

Answer:

a. The probability that a randomly - selected male college student gains between \(0\) kg and \(3\) kg is \(0.2434\).
b. The probability that the mean weight gain of \(9\) randomly - selected male college students is between \(0\) kg and \(3\) kg is \(0.6505\).
c. C. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.