Trevor earns $7.80 an hour take home pay and works 20 hours each week. Trevor wants to buy a new laptop computer to take to coll
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Answer:
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!