SOLUTION: A distribution of values is normal with a mean of 210 and a standard deviation of 23. From this distribution, you are drawing samples of size 33. Find the interval containing th

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Question 1119447: A distribution of values is normal with a mean of 210 and a standard deviation of 23. From this distribution, you are drawing samples of size 33.
Find the interval containing the middle-most 94% of sample means:

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
mean is 210
standard deviation is 23
sample size is 33.

standard error = standard deviation divided by square root of sample size = sqrt(23/33) = .8348471099.

you want the middle 94% of the distribution of sample means.

that would be .94 of the area under the normal distribution curve.

1 - .94 = .6 area that is split between the lower end and the upper end.

that makes .03 area on the left of the confidence interval and .03 area on the right of the confidence interval.

an area of .03 to the left of the confidence interval is associated with a z-score of -1.88079361.

since the normal distribution curve is symmetric about the mean, then your confidence interval is between a z-score of plus or minus 1.88079361.

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 raw mean
s is the standard error.

you get 1.88079361 = (x-210)/.8348471099 and you get -1.88079361 = (x-210)/.8348471099.

solve for x to get:

x = .8348471099 * 1.88079361 + 210 and you get x = .8348471099 * -1.88079361 + 210.

this results in a confidence interval that is 94% of the area under the normal distribution curve that goes between the raw score of 208.4298249 to 211.5701751.

this can be seen in the following z-score calculator output.

the first tells you the critical z-score for an area of .94 in the middle of the normal distribution curve.

the second tells you the critical raw score for an area of.94 in the middle of the normal distribution curve.

the first has a mean of 0 and a standard deviation of 1 which is the mean and standard deviation for the z-score.

the second has a mean of 210 and a standard deviation (actually a standard error) of .8348471099.

there is some rounding of the inputs when using this calculator.

this is more than enough for most applications, especially when the final answer is normally rounded to 3 decimal digits.

here's the display of the calculations.

$$$

$$$









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