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Let's go through each question step by step:

### **Question 3:**
Given that \( x \) is a normally distributed variable with a mean of \( 60 \) and a standard deviation of \( 10 \), find the following probabilities:

#### a) \( P(x > 60) \)

Since \( x \) follows a normal distribution, the probability of \( x \) being equal to the mean (60) is exactly 0.5. The area to the right of 60 in a standard normal distribution (which has a mean of 0) is also 0.5.

So:
\[
P(x > 60) = 0.5
\]

#### b) \( P(x < 72) \)

First, we calculate the z-score for 72:
\[
z = \frac{72 - 60}{10} = 1.2
\]

We then use the standard normal distribution table to find \( P(z < 1.2) \). The value for \( z = 1.2 \) is approximately 0.8849.

Thus:
\[
P(x < 72) = 0.8849
\]

#### c) \( P(65 < x < 82) \)

We need to find two z-scores:

1. For \( x = 65 \):
\[
z_1 = \frac{65 - 60}{10} = 0.5
\]
From the standard normal table, \( P(z < 0.5) \approx 0.6915 \).

2. For \( x = 82 \):
\[
z_2 = \frac{82 - 60}{10} = 2.2
\]
From the table, \( P(z < 2.2) \approx 0.9861 \).

Now, to find \( P(65 < x < 82) \):
\[
P(65 < x < 82) = P(z < 2.2) - P(z < 0.5) = 0.9861 - 0.6915 = 0.2946
\]

### **Question 4:**
The waiting time \( x \) for a lecture to begin follows a normal distribution with a mean of \( 3.7 \) minutes and a standard deviation of \( 1.4 \) minutes.

#### a) Probability that a randomly selected course has to wait for less than 2.0 minutes.

Calculate the z-score for 2.0:
\[
z = \frac{2.0 - 3.7}{1.4} = -1.21
\]

Using the standard normal table, \( P(z < -1.21) \approx 0.1131 \).

So:
\[
P(x < 2.0) = 0.1131
\]

#### b) Probability that a randomly selected course has to wait between 5.0 and 7.5 minutes.

1. For \( x = 5.0 \):
\[
z_1 = \frac{5.0 - 3.7}{1.4} = 0.93
\]
From the table, \( P(z < 0.93) \approx 0.8238 \).

2. For \( x = 7.5 \):
\[
z_2 = \frac{7.5 - 3.7}{1.4} = 2.71
\]
From the table, \( P(z < 2.71) \approx 0.9966 \).

Now, to find \( P(5.0 < x < 7.5) \):
\[
P(5.0 < x < 7.5) = P(z < 2.71) - P(z < 0.93) = 0.9966 - 0.8238 = 0.1728
\]

### Summary:
- **3a)** \( P(x > 60) = 0.5 \)
- **3b)** \( P(x < 72) = 0.8849 \)
- **3c)** \( P(65 < x < 82) = 0.2946 \)
- **4a)** \( P(x < 2.0) = 0.1131 \)
- **4b)** \( P(5.0 < x < 7.5) = 0.1728 \)

Would you like further explanations or details on any of these parts? 

Here are 5 related questions for further exploration:

1. How do you calculate the z-score for a given value in a normal distribution?
2. What is the significance of the standard normal distribution table in solving these problems?
3. How do you interpret probabilities in a normal distribution?
4. How would the probabilities change if the standard deviation was smaller or larger?
5. What happens to the probability \( P(x > a) \) as \( a \) increases beyond the mean?

**Tip:** Always visualize the normal distribution curve when solving probability questions; it helps to understand the area under the curve corresponding to the probabilities you seek.

Given that x is a normally distributed variable with a mean of 60 and a standard deviation of 10, find P(x > 60), P(x < 72), P(65 < x < 82); and the waiting time x for a lecture is approximately normally distributed with a mean of 3.7 minutes and standard deviation of 1.4 minutes, find the probability that a randomly selected course has to wait for less than 2.0 minutes, and the probability that a course has to wait between 5.0 and 7.5 minutes.

Learn how to calculate probabilities using the normal distribution formula. Example includes finding P(x > 60), P(x < 72), and P(65 < x < 82) for a variable with mean 60 and standard deviation 10. Also, determine the waiting time for a lecture based on a normal distribution with a mean of 3.7 minutes and a standard deviation of 1.4 minutes.

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