The average production cost for major movies is 65 million dollars and the standard deviation is 18 million dollars. Assume the production cost distribution is normal. Suppose that 39 randomly selected major movies are researched. Answer the following questions. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X~ N(,)b. What is the distribution of Xbar?Xbar~N(,)c. For a single randomly selected movie, find the probability that this movie's production cost is between 66 and 69 million dollars. d. For the aroup of 39 movies, find the probability that the average production cost is between 66 and 69 milion dollars e. For part d), is the assumption of normal necessary? a) No b) Yes

Step-by-step explanation:

Hello!

The variable of interest is

X: production cost of a major move

Its average is μ= 65 million dollars, and standard deviation σ= 18 million dollars.

a)

This variable has a normal distribution X~N(μ;σ²)

X~N(65;324)

b)

The distribution of the sample mean has the same shape as the distribution of the variable, but its variance is affected by the sample size:

~N(μ;σ²/n) ⇒ ~N(65;8.3077)

σ²/n= 324/39= 8.30769≅ 8.3077

c)

You have to calculate the probability of a single movie costing between 69 and  66 million dollars, symbolically:

P(66≤X≤69)= (X≤69)-P(X≤66)

You have to use the standard normal distribution to calculate this probability, so first you have to calculate the Z values that correspond to each value of X using: Z= (X-μ)/σ  ~ N(0;1)

Z₁= (69-65)/18= 0.22

Z₂=(66-65)/18= 0.05

Now you look for the corresponding probability values using the standard normal table

P(Z≤0.22)= 0.58706

P(Z≤0.05)= 0.51994

P(66≤X≤69)= (X≤69)-P(X≤66)

P(Z≤0.22)-P(Z≤0.05)= 0.58706 - 0.51994= 0.06712

d)

Now you have to calculate the probability of the average production cost to be between 69 and 66 million. Since the probability is for the average value of the sample, you have to work using the distribution of the sample mean. The values od Z are to be calculated using the formula Z=  (-μ)/(σ/√n)

σ/√n= 2.8823

P(66≤≤69)= (≤69)-P(≤66)

Z₁= (69-65)/2.8823= 1.387= 1.39

Z₂= (66-65)/2.8823= 0.346= 0.35

P(Z≤1.39)-P(Z≤0.35)= 0.91774 - 0.63683= 0.28091

e)

No

If the variable has an unknown or non-normal distribution, but the sample is large enough (normally a sample n≥30 is considered to be large) you can apply the central limit theorem and approximate the sampling distribution to normal.

I hope this helps!

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %

2x=70  

x=70/2

x=35

thats the answer

step-by-step explanation: