SOLUTION: Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The middle 90% of the heights are between what two v

Question 1196282: Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The middle 90% of the heights are between what two values? Round to one decimal place.

60.4 inches and 66.8 inches

59.5 inches and 67.7 inches

63.1 inches and 64.1 inches

61.9 inches and 65.3 inches

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

mu = 63.6 = population mean
sigma = 2.5 = population standard deviation

Since we're dealing with the middle 90%, each tail has an area of (1 - 0.9)/2 = 0.05
Note that 0.05 + 0.90 + 0.05 = 1
The area under the entire curve is always 1.

At 90% confidence, the z critical value is about 1.645
Use a table like this to determine the critical value
https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
Look blue row at the bottom marked "Z"
The value 1.645 is just above the 90% confidence level

This means P(-1.645 < Z < 1.645) = 0.90 approximately
Roughly 90% of the area under the standard normal curve is between -1.645 and 1.645

Let's find the raw score x that leads to z = -1.645
z = (x - mu)/sigma
-1.645 = (x - 63.6)/2.5
-1.645*2.5 = x - 63.6
-4.1125 = x - 63.6
x - 63.6 = -4.1125
x = -4.1125 + 63.6
x = 59.4875
x = 59.5
This is the approximate lower bound

Now repeat those steps with z = 1.645
z = (x - mu)/sigma
1.645 = (x - 63.6)/2.5
1.645*2.5 = x - 63.6
4.1125 = x - 63.6
x - 63.6 = 4.1125
x = 4.1125 + 63.6
x = 67.7125
x = 67.7
This is the approximate upper bound

Therefore, the middle 90% is between 59.5 inches and 67.7 inches which is choice B

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