These are the answers in red. Just need to see how my professor received the ans

Question

These are the answers in red. Just need to see how my professor received the answers. Please help


These are the answers in red. Just need to see how my professor received the answers. Please help
2 9) The distribution of the number of barrels of oil produced each day by a certain oil well is normally distributed with a mean of 400 and a standard deviation of 70. a) If a sample of 40 days were taken, what is the probability the mean number of barrels of oil produced would be between 380 and 410? (3 points) 0.7808 b) If statsitics for any individual day were selected, what is the probability the number of barrels of oil produced would be between 380 and 410? (3 points) 0.1698

Explanation / Answer

a.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 400
standard Deviation ( sd )= 70/ Sqrt ( 40 ) =11.068
sample size (n) = 40
a.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 380) = (380-400)/70/ Sqrt ( 40 )
= -20/11.06797
= -1.80702
= P ( Z <-1.80702) From Standard Normal Table
= 0.03538
P(X < 410) = (410-400)/70/ Sqrt ( 40 )
= 10/11.06797 = 0.90351
= P ( Z <0.90351) From Standard Normal Table
= 0.81687
P(380 < X < 410) = 0.81687-0.03538 = 0.7808
b.
mean ( u ) = 400
standard Deviation ( sd )= 70
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 380) = (380-400)/70
= -20/70 = -0.2857
= P ( Z <-0.2857) From Standard Normal Table
= 0.3875
P(X < 410) = (410-400)/70
= 10/70 = 0.1429
= P ( Z <0.1429) From Standard Normal Table
= 0.5568
P(380 < X < 410) = 0.5568-0.3875 = 0.1698

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