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Let's go through the problems in the image step by step.

### Problem a: The distribution of X
We are given that the speed of cars on the freeway is normally distributed with a mean of 75 mph and a standard deviation of 8 mph. The distribution of X (the speed of a randomly selected car) is written as:

\[
X \sim N(\mu, \sigma^2)
\]

Here:
- \(\mu = 75\) (mean)
- \(\sigma = 8\) (standard deviation)

So, the distribution is:

\[
X \sim N(75, 8^2) \quad \text{or} \quad X \sim N(75, 64)
\]

### Problem b: Probability that a car is traveling more than 72 mph
We need to find \( P(X > 72) \).

To do this, we first standardize the value by finding the z-score:
\[
z = \frac{X - \mu}{\sigma} = \frac{72 - 75}{8} = \frac{-3}{8} = -0.375
\]

Now, we look up the z-score in a standard normal distribution table or use a calculator to find \( P(Z > -0.375) \).

From the table:
\[
P(Z > -0.375) = 0.6462
\]

Thus, the probability that a car is traveling more than 72 mph is approximately **0.6462**.

### Problem c: Probability that a car is traveling between 74 mph and 78 mph
We need to find \( P(74 < X < 78) \).

First, we calculate the z-scores for both 74 mph and 78 mph:
- For 74 mph: 
  \[
  z = \frac{74 - 75}{8} = \frac{-1}{8} = -0.125
  \]
  
- For 78 mph:
  \[
  z = \frac{78 - 75}{8} = \frac{3}{8} = 0.375
  \]

Using the standard normal distribution table:
- \( P(Z < -0.125) = 0.4503 \)
- \( P(Z < 0.375) = 0.6462 \)

Now, the probability that the car is traveling between 74 mph and 78 mph is:
\[
P(74 < X < 78) = P(Z < 0.375) - P(Z < -0.125) = 0.6462 - 0.4503 = 0.1959
\]

Thus, the probability is approximately **0.1959**.

### Problem d: 90% of cars travel at least how fast on the freeway?
We need to find the speed at which 90% of cars are traveling at least that fast, i.e., the 10th percentile of the distribution.

Using the z-table, the z-score corresponding to the 10th percentile is approximately \(-1.2816\).

Now, we convert this z-score back to the original units using the mean and standard deviation:
\[
X = \mu + z\sigma = 75 + (-1.2816)(8) = 75 - 10.2528 = 64.7472
\]

Thus, 90% of all cars travel at least **64.7472 mph**.

---

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

### Related Questions:
1. How would you calculate the probability that a car is traveling more than 80 mph?
2. What is the probability that a car is traveling less than 70 mph?
3. How would the results change if the standard deviation was 10 mph instead of 8 mph?
4. How can we use the normal distribution to find the median speed of cars on the freeway?
5. What is the probability that a randomly chosen car is traveling exactly at the mean speed?

### Tip:
Always convert values to z-scores when working with probabilities involving the normal distribution.

Suppose that the speed at which cars go on the freeway is normally distributed with a mean 75 mph and standard deviation 8 miles per hour. Let X be the speed for a randomly selected car. Round all answers to 4 decimal places where possible.

a. What is the distribution of X?
b. If one car is randomly chosen, find the probability that it is traveling more than 72 mph.
c. If one of the cars is randomly chosen, find the probability that it is traveling between 74 and 78 mph.
d. 90% of all cars travel at least how fast on the freeway?

Solve a probability problem involving speeds on the freeway. Given that car speeds are normally distributed with a mean of 75 mph and a standard deviation of 8 mph, find the probability of a car traveling more than 72 mph, between 74 and 78 mph, and calculate the 10th percentile speed.

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