Question 1198038: A population of values has a normal distribution with μ = 76.3 and σ = 41.4 . If a random sample of size n = 22 is selected,
Find the probability that a single randomly selected value is less than 77.2. Round your answer to four decimals.
P(X < 77.2) =
Find the probability that a sample of size n = 22 is randomly selected with a mean less than 77.2. Round your answer to four decimals.
P(M < 77.2) =
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! population mean = 76.3
population standard deviation = 41.4
sample size = 22
problem 1:
when you are looking for a randomly selected element, then the standard deviation is used.
z = (x-m)/s
z is the z-score
x is the value of the random sample element value
m is the value of the population mean
s is the standard deviation.
the formula becomes:
z = (77.2 - 76.3) / 41.4 = .021739 rounded to 6 decimal places.
p(z < .021739) = area to the left of that z-score = .5087 rounded to 4 decimal places.
problem 2:
when you are looking for the mean of a sample, then the standard error is used.
standard error = standard deviation / square root of sample size = 41.4 / square root of 22 = 8.8265 rounded to 4 decimal places.
the same z-score formula is used.
z = (77.2 - 76.3) / 8.8265 = .101966 rounded to 6 decimal places.
p(z < .101966) = .5406 rounded to 4 decimal places.
the first problem is looking at a distribution of randomly selected population elements.
the second problem is looking at a distribution of randomly selected means of samples of size 22.
as the sample size gets larger, the distribution of the mean of samples becomes tighter, which explains why the formula for the standard error is equal to the standard deviation divided by the square root of the sample size.
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