Question

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

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

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

Answer 1

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**