Question

The average production cost for major movies is 65 million dollars and the standard deviation is 19 million dollars. Assume the production cost distribution is normal. Suppose that 46 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.

1. What is the distribution of X?X ~ N(_______________,______________)
2. What is the distribution of ¯x ? ¯x ~ N(_____________,______________)
3. For a single randomly selected movie, find the probability that this movie's production cost is between 64 and 70 million dollars.
4. For the group of 46 movies, find the probability that the average production cost is between 64 and 70 million dollars.
5. For part d), is the assumption of normal necessary? No Yes

Answer 1

ANSWER:

Given that,

The average production cost for major movies is 65 million dollars and the standard deviation is 19 million dollars. Assume the production cost distribution is normal. Suppose that 46 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.

Mean = = 65 million dollars

Standard deviation = = 19 million dollars

Sample size = n = 46

(a) What is the distribution of X?

X N ( , )

X N ( 65 , 19)

(b) What is the distribution of ¯x ?

N ( , )

N ( , /sqrt(n))

= = 65

= /sqrt(n) = 19/sqrt(46) = 2.8014

N ( 65 , 2.8014)

(c) For a single randomly selected movie, find the probability that this movie's production cost is between 64 and 70 million dollars.

P(64 < x < 70) = P((64-65)/19 < (x-)/ < (70-65)/19)

P(64 < x < 70) = P(-1 /19 < z < 5/19)

P(64 < x < 70) = P(-0.05 < z < 0.26)

P(64 < x < 70) = P(z < 0.26)-P(z < -0.05)

From z score table as given below

P(64 < x < 70) = 0.60257-0.48006

P(64 < x < 70) = 0.1225

(d) For the group of 46 movies, find the probability that the average production cost is between 64 and 70 million dollars.

P(64 < < 70) = P((64-65)/(19/sqrt(46)) < (-)/(/sqrt(n) < (70-65)/(19/sqrt(46)))

P(64 < < 70) = P(-1 /(19/sqrt(46))< z < 5/(19/sqrt(46)))

P(64 < < 70) = P(-0.37 < z < 1.78)

P(64 < < 70) = P(z < 1.78)-P(z < -0.37)

From z score table as given below

P(64 < < 70) = 0.96246-0.35569

P(64 < < 70) = 0.6068

(e) For part (d) is the assumption of normal necessary ?

No

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