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The average production cost for major movies is 57 million dollars and the standard deviation is 22 million dollars. Assume the production cost distribution is normal. Suppose that 17 randomly selected major movies are researched. Answer the following questions. Give your answers in millions of dollars, not dollars. Round all answers to 4 decimal places where possible. What is the distribution of X ? X ~ N( , ) What is the distribution of ¯ x ? ¯ x ~ N( , ) For a single randomly selected movie, find the probability that this movie's production cost is between 55 and 58 million dollars. For the group of 17 movies, find the probability that the average production cost is between 55 and 58 million dollars. For part d), is the assumption of normal necessary? NoYes

Sagot :

Using the normal distribution, we have that:

  • The distribution of X is [tex]X \approx (57,22)[/tex].
  • The distribution of [tex]\mathbf{\bar{X}}[/tex] is [tex]\bar{X} \approx (57, 5.3358)[/tex].
  • 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
  • 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem, the parameters are given as follows:

[tex]\mu = 57, \sigma = 22, n = 17, s = \frac{22}{\sqrt{17}} = 5.3358[/tex]

Hence:

  • The distribution of X is [tex]X \approx (57,22)[/tex].
  • The distribution of [tex]\mathbf{\bar{X}}[/tex] is [tex]\bar{X} \approx (57, 5.3358)[/tex].

The probabilities are the p-value of Z when X = 58 subtracted by the p-value of Z when X = 55, hence, for a single movie:

X = 58:

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

[tex]Z = \frac{58 - 57}{22}[/tex]

Z = 0.05.

Z = 0.05 has a p-value of 0.5199.

X = 55:

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

[tex]Z = \frac{55 - 57}{22}[/tex]

Z = -0.1.

Z = -0.1 has a p-value of 0.4602.

0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.

For the sample of 17 movies, we have that:

X = 58:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{58 - 57}{5.3358}[/tex]

Z = 0.19.

Z = 0.19 has a p-value of 0.5753.

X = 55:

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{55 - 57}{5.3358}[/tex]

Z = -0.38.

Z = -0.38 has a p-value of 0.3520.

0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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