Answer :
Certainly! Let's walk through the steps to find the probability that a randomly selected male college student gains between \(0 \text{ kg}\) and \(3 \text{ kg}\) during his freshman year, given that the weight gain is normally distributed with a mean \(\mu = 1.3 \text{ kg}\) and a standard deviation \(\sigma = 5.3 \text{ kg}\).
1. Identify the parameters:
- Mean (\(\mu\)): \(1.3 \text{ kg}\)
- Standard deviation (\(\sigma\)): \(5.3 \text{ kg}\)
- Lower bound of weight gain: \(0 \text{ kg}\)
- Upper bound of weight gain: \(3 \text{ kg}\)
2. Calculate the z-scores for the lower and upper bounds:
The \(z\)-score is calculated as:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
- For the lower bound \(0 \text{ kg}\):
[tex]\[ z_{\text{lower}} = \frac{0 - 1.3}{5.3} \approx -0.245283 \][/tex]
- For the upper bound \(3 \text{ kg}\):
[tex]\[ z_{\text{upper}} = \frac{3 - 1.3}{5.3} \approx 0.320755 \][/tex]
3. Use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities corresponding to these z-scores:
- The probability corresponding to \(z_{\text{lower}} \approx -0.245283\) is:
[tex]\[ P(Z \leq -0.245283) \][/tex]
- The probability corresponding to \(z_{\text{upper}} \approx 0.320755\) is:
[tex]\[ P(Z \leq 0.320755) \][/tex]
4. Calculate the probability that the weight gain is between \(0 \text{ kg}\) and \(3 \text{ kg}\):
The probability is found by taking the difference between the probabilities for the upper bound and lower bound:
[tex]\[ P(0 \leq X \leq 3) = P(Z \leq 0.320755) - P(Z \leq -0.245283) \approx 0.222683 \][/tex]
So, the probability that a randomly selected male college student gains between [tex]\(0 \text{ kg}\)[/tex] and [tex]\(3 \text{ kg}\)[/tex] during his freshman year is approximately [tex]\(0.2227\)[/tex].
1. Identify the parameters:
- Mean (\(\mu\)): \(1.3 \text{ kg}\)
- Standard deviation (\(\sigma\)): \(5.3 \text{ kg}\)
- Lower bound of weight gain: \(0 \text{ kg}\)
- Upper bound of weight gain: \(3 \text{ kg}\)
2. Calculate the z-scores for the lower and upper bounds:
The \(z\)-score is calculated as:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
- For the lower bound \(0 \text{ kg}\):
[tex]\[ z_{\text{lower}} = \frac{0 - 1.3}{5.3} \approx -0.245283 \][/tex]
- For the upper bound \(3 \text{ kg}\):
[tex]\[ z_{\text{upper}} = \frac{3 - 1.3}{5.3} \approx 0.320755 \][/tex]
3. Use the cumulative distribution function (CDF) of the standard normal distribution to find the probabilities corresponding to these z-scores:
- The probability corresponding to \(z_{\text{lower}} \approx -0.245283\) is:
[tex]\[ P(Z \leq -0.245283) \][/tex]
- The probability corresponding to \(z_{\text{upper}} \approx 0.320755\) is:
[tex]\[ P(Z \leq 0.320755) \][/tex]
4. Calculate the probability that the weight gain is between \(0 \text{ kg}\) and \(3 \text{ kg}\):
The probability is found by taking the difference between the probabilities for the upper bound and lower bound:
[tex]\[ P(0 \leq X \leq 3) = P(Z \leq 0.320755) - P(Z \leq -0.245283) \approx 0.222683 \][/tex]
So, the probability that a randomly selected male college student gains between [tex]\(0 \text{ kg}\)[/tex] and [tex]\(3 \text{ kg}\)[/tex] during his freshman year is approximately [tex]\(0.2227\)[/tex].
