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
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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.
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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