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Let's go through the questions step by step, using the information from the problem:

### Given:
- Population mean \(\mu = 59\) million dollars
- Population standard deviation \(\sigma = 22\) million dollars
- Sample size \(n = 48\)

#### Part (a): Distribution of \(X\)
The problem states that the production cost distribution is normal. For a single movie, \(X\) represents the production cost.

So, the distribution of \(X\) is:
\[
X \sim N(59, 22^2)
\]
This is because the normal distribution is characterized by the mean \(\mu = 59\) and the variance \(\sigma^2 = 22^2\).

#### Part (b): Distribution of \(\bar{x}\)
For the sample of 48 movies, the sampling distribution of the sample mean \(\bar{x}\) will also be normal, but with a reduced standard deviation given by:
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{22}{\sqrt{48}} \approx 3.1735
\]
So, the distribution of \(\bar{x}\) is:
\[
\bar{x} \sim N(59, 3.1735^2)
\]

#### Part (c): Probability that a single movie's production cost is between 53 and 59 million dollars
To find this probability, we will standardize the values and use the standard normal distribution (Z-distribution).

For \(X = 53\) million:
\[
Z = \frac{53 - 59}{22} = \frac{-6}{22} \approx -0.2727
\]

For \(X = 59\) million:
\[
Z = \frac{59 - 59}{22} = 0
\]

Using the Z-table, the probability of \(Z\) being between \(-0.2727\) and 0 is about 0.1075.

Thus, the probability is approximately **0.1075**.

#### Part (d): Probability that the average production cost for the group of 48 movies is between 53 and 59 million dollars
We now use the sampling distribution of \(\bar{x}\), which has a standard deviation of 3.1735.

For \(\bar{x} = 53\) million:
\[
Z = \frac{53 - 59}{3.1735} = \frac{-6}{3.1735} \approx -1.8904
\]

For \(\bar{x} = 59\) million:
\[
Z = \frac{59 - 59}{3.1735} = 0
\]

Using the Z-table, the probability of \(Z\) being between \(-1.8904\) and 0 is approximately 0.4706.

Thus, the probability is approximately **0.4706**.

#### Part (e): Is the assumption of normality necessary for part (d)?
Since the sample size \(n = 48\) is sufficiently large (greater than 30), by the **Central Limit Theorem**, the sampling distribution of the sample mean \(\bar{x}\) will be approximately normal even if the original distribution of \(X\) were not normal. 

Thus, **no**, the assumption of normality is not strictly necessary for part (d), as the Central Limit Theorem justifies the approximation.

---

Would you like more details on any part or clarification? Here are five related questions to expand your understanding:

1. How would the results change if the sample size were smaller (e.g., \(n = 10\))?
2. What is the impact of increasing the standard deviation on the probabilities calculated?
3. How does the Central Limit Theorem apply in situations with non-normal distributions?
4. Could we calculate the same probabilities using a t-distribution instead of a normal distribution?
5. What if the range for probability calculations was shifted to 60 and 70 million dollars?

**Tip:** When working with sample means, always check if the sample size is large enough to apply the Central Limit Theorem for normal approximation.

The average production cost for major movies is 59 million dollars and the standard deviation is 22 million dollars. Assume the production cost distribution is normal. Suppose that 48 randomly selected major movies are researched. Answer the following questions.
(a) What is the distribution of X?
(b) What is the distribution of x̄?
(c) For a single randomly selected movie, find the probability that this movie's production cost is between 53 and 59 million dollars.
(d) For the group of 48 movies, find the probability that the average production cost is between 53 and 59 million dollars.
(e) Is the assumption of normal necessary for part d?

Solve a problem involving movie production costs using normal distribution. This question includes the probability of costs within certain ranges and uses concepts like the sampling distribution, Z-scores, and the Central Limit Theorem. Suitable for college-level statistics.

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