SOLUTION: A population of values has a normal distribution with μ=225.8 and σ=87.4. You intend to draw a random sample of size n=215.
Find P82, which is the mean separating the
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-> SOLUTION: A population of values has a normal distribution with μ=225.8 and σ=87.4. You intend to draw a random sample of size n=215.
Find P82, which is the mean separating the
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Question 1119446: A population of values has a normal distribution with μ=225.8 and σ=87.4. You intend to draw a random sample of size n=215.
Find P82, which is the mean separating the bottom 82% means from the top 18% means. Answer by Theo(13342) (Show Source):
standard error = standard deviation divided by square root of sample size.
standard error = 87.4 / sqrt(215) = 5.960630527.
you look up in the normal distribution table, or use a normal distribution calculator, to find the z-score that is associated with 829% of the area under the normal distribution curve that is to the left of it.
i used the normal distribution calculator that is in the TI-84 Plus and came up with a z-score of .915365082.
to find the raw score associated with that, use the z-score formula of:
z = (x-m)/s
z is the z-score.
x is the raw score.
m is the mean
s is the standard error.
that formula becomes:
.915365082 = (x-225.8) / 5.960630527.
solve for x to get:
x = .915365082 * 5.960630527 + 225.8.
that results in x = 231.2561531.
that score will have 82% of the normal distribution curve to the left of it and 18% of the normal distribution curve to the right of it.
that means the probability of getting a raw score less than 231.2561531 is 82% and the probability of getting a raw score greater than 231.2561531 is 18%.
visually, this looks like this:
the first display shows you the proability of getting a raw score less than 231.2561.
the second display shows you the probability of getting a raw score greater than 231.2561.
round your answer as required.
any difference between what this online calculator shows you and the calculator i used has to do how many decimal places rounding has occurred.