Questions: Assume the amounts of weight that male college students gain during their freshman year are normally distributed with a mean of μ=1.2 kg and a standard deviation of σ=4.9 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 0.2401. (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 0.6316. (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 distribution is of individuals, not sample means, the distribution is a normal distribution for any sample size. B. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size. C. Since the weight gain exceeds 30, the distribution of sample means 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.

Solution Steps

To find the probability that a randomly selected male college student gains between \(0 \, \text{kg}\) and \(3 \, \text{kg}\), we first calculate the Z-scores for the given values using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

where:

• \(X\) is the value,
• \(\mu = 1.2 \, \text{kg}\) (mean),
• \(\sigma = 4.9 \, \text{kg}\) (standard deviation).

Calculating the Z-scores:

• For \(X = 3\): \[ Z_{end} = \frac{3 - 1.2}{4.9} \approx 0.3673 \]
• For \(X = 0\): \[ Z_{start} = \frac{0 - 1.2}{4.9} \approx -0.2449 \]

Using the cumulative distribution function \( \Phi \), we find: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.3673) - \Phi(-0.2449) \approx 0.2401 \]

Thus, the probability that one student gains between \(0 \, \text{kg}\) and \(3 \, \text{kg}\) is \(0.2401\).

Next, we calculate the probability that the mean weight gain of \(9\) male college students is between \(0 \, \text{kg}\) and \(3 \, \text{kg}\). For this, we use the standard error of the mean, which is given by:

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.9}{\sqrt{9}} = \frac{4.9}{3} \approx 1.6333 \]

Calculating the Z-scores for the mean:

• For \(X = 3\): \[ Z_{end} = \frac{3 - 1.2}{1.6333} \approx 1.102 \]
• For \(X = 0\): \[ Z_{start} = \frac{0 - 1.2}{1.6333} \approx -0.7347 \]

Using the cumulative distribution function \( \Phi \), we find: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.102) - \Phi(-0.7347) \approx 0.6335 \]

Thus, the probability that the mean weight gain of \(9\) students is between \(0 \, \text{kg}\) and \(3 \, \text{kg}\) is \(0.6335\).

The reason we can use the normal distribution in part (b), even though the sample size does not exceed \(30\), is because the original population has a normal distribution. Therefore, the distribution of sample means is also a normal distribution for any sample size.

Final Answer