Question

question 3
the average production cost for major movies is 66 million dollars and the standard deviation is 19 million dollars. assume the production cost distribution is normal. suppose that 40 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.
a. what is the distribution of x? x ~ n(66,000,000,19,000,000)
b. what is the distribution of \\(\bar{x}\\)? \\(\bar{x}\\) ~ n(66,000,000,)
c. for a single randomly selected movie, find the probability that this movies production cost is between 63 and 67 million dollars.
d. for the group of 40 movies, find the probability that the average production cost is between 63 and 67 million dollars.
e. for part d), is the assumption of normal necessary? no yes
hint:
some helpful videos:

• finding the sampling distribution +
• finding a probability using the central limit theorem +
• finding value given a probability using the central limit theorem +
• the central limit theorem for sums +

Explanation:

Step1: Recall normal - distribution properties for part a

The random variable \(X\) representing the production cost of a single major movie follows a normal distribution with mean \(\mu = 66\) million dollars and standard deviation \(\sigma=19\) million dollars. So \(X\sim N(66,19)\).

Step2: Recall sampling - distribution of the sample mean for part b

The sampling distribution of the sample mean \(\bar{X}\) of a sample of size \(n = 40\) from a normal population has mean \(\mu_{\bar{X}}=\mu = 66\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{19}{\sqrt{40}}\approx 3.0092\). So \(\bar{X}\sim N(66,\frac{19}{\sqrt{40}})\approx N(66,3.0092)\).

Step3: Standardize for part c

For a single movie, we standardize \(x_1 = 63\) and \(x_2 = 67\) using \(z=\frac{x-\mu}{\sigma}\).
For \(x = 63\), \(z_1=\frac{63 - 66}{19}=\frac{- 3}{19}\approx - 0.1579\).
For \(x = 67\), \(z_2=\frac{67 - 66}{19}=\frac{1}{19}\approx0.0526\).
Then \(P(63<X<67)=P(-0.1579<Z<0.0526)=\Phi(0.0526)-\Phi(-0.1579)\), where \(\Phi(z)\) is the cumulative - distribution function of the standard normal distribution.
\(\Phi(0.0526)\approx0.5210\), \(\Phi(-0.1579)=1 - \Phi(0.1579)\approx1 - 0.5636 = 0.4364\).
So \(P(63<X<67)=0.5210 - 0.4364=0.0846\).

Step4: Standardize for part d

For the sample of \(n = 40\) movies, we standardize \(\bar{x}_1 = 63\) and \(\bar{x}_2 = 67\) using \(z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}\).
\(z_1=\frac{63 - 66}{3.0092}=\frac{-3}{3.0092}\approx - 0.9969\).
\(z_2=\frac{67 - 66}{3.0092}=\frac{1}{3.0092}\approx0.3323\).
\(P(63<\bar{X}<67)=P(-0.9969<Z<0.3323)=\Phi(0.3323)-\Phi(-0.9969)\).
\(\Phi(0.3323)\approx0.6293\), \(\Phi(-0.9969)=1 - \Phi(0.9969)\approx1 - 0.8400 = 0.1600\).
So \(P(63<\bar{X}<67)=0.6293 - 0.1600 = 0.4693\).

Step5: Apply Central - Limit Theorem for part e

By the Central - Limit Theorem, when \(n = 40\) (\(n\geq30\)), the sampling distribution of the sample mean \(\bar{X}\) is approximately normal regardless of the shape of the population distribution. So the answer is No.

Answer:

a. \(X\sim N(66,19)\)
b. \(\bar{X}\sim N(66,\frac{19}{\sqrt{40}})\approx N(66,3.0092)\)
c. \(0.0846\)
d. \(0.4693\)
e. No