A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 27 and standard deviation 5. Round to 4 decimal places.

To solve these problems, we will use the properties of the normal distribution. For a normally distributed variable \( X \) with mean \( \mu = 27 \) and standard deviation \( \sigma = 5 \), we can standardize our values using the Z-score formula:

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

where:

• \( X \) is the value we are interested in,
• \( \mu \) is the mean,
• \( \sigma \) is the standard deviation.

Then, we'll use the standard normal distribution table or a calculator to find the required probabilities.

(a) What proportion of students studied more than 39 hours?

1. Calculate the Z-score for 39 hours: \[ Z = \frac{39 - 27}{5} = \frac{12}{5} = 2.4 \]

2. Find the cumulative probability \( P(Z < 2.4) \) using a standard normal distribution table or calculator. From the Z-table, \( P(Z < 2.4) \approx 0.9918 \).

3. Calculate the proportion of students studying more than 39 hours: \[ P(X > 39) = 1 - P(Z < 2.4) = 1 - 0.9918 = 0.0082. \]

So, the proportion of students who studied more than 39 hours is approximately 0.0082.

(b) What is the probability that a randomly selected student spent between 14 and 33 hours studying?

1. Calculate the Z-score for 14 hours: \[ Z = \frac{14 - 27}{5} = \frac{-13}{5} = -2.6 \]

   Using the Z-table, \( P(Z < -2.6) \approx 0.0047 \).

2. Calculate the Z-score for 33 hours: \[ Z = \frac{33 - 27}{5} = \frac{6}{5} = 1.2 \]

   Using the Z-table, \( P(Z < 1.2) \approx 0.8849 \).

3. Calculate the probability that a student spent between 14 and 33 hours: \[ P(14 < X < 33) = P(Z < 1.2) - P(Z < -2.6) = 0.8849 - 0.0047 = 0.8802. \]

So, the probability that a randomly selected student spent between 14 and 33 hours studying is approximately 0.8802.

(c) What proportion of students studied less than 33 hours?

1. From the previous calculation, we already found \( P(Z < 1.2) \) for 33 hours: \[ P(X < 33) = P(Z < 1.2) \approx 0.8849. \]

So, the proportion of students who studied less than 33 hours is approximately 0.8849.

Summary of Answers:

• (a) Proportion studied more than 39 hours: 0.0082
• (b) Probability spent between 14 and 33 hours: 0.8802
• (c) Proportion studied less than 33 hours: 0.8849