SOLUTION: A city in the USA is trying to increase tourism. The tourism division would like to include in its tourism brochure the middle range of temperatures that occur on 90% of the days

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Question 40009: A city in the USA is trying to increase tourism. The tourism division would like to include in its tourism brochure the middle range of temperatures that occur on 90% of the days there. A check with the Weather Service finds that the average temperature in this city is 76 degrees with a standard deviation of 5.7 (the temperatures were normally distributed). Find the two temperatures that cut off the middle 90% of temperatures in this city.
I don't know the source of this problem. I'm trying to get some help for a friend who is trying to understand these type of problems. Thank you for your assistance.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A city in the USA is trying to increase tourism. The tourism division would like to include in its tourism brochure the middle range of temperatures that occur on 90% of the days there. A check with the Weather Service finds that the average temperature in this city is 76 degrees with a standard deviation of 5.7 (the temperatures were normally distributed). Find the two temperatures that cut off the middle 90% of temperatures in this city.
Sketch a normal curve. Put mean at center = 76 and note that standard dev.
is 5.7
Sketch a standard normal curve with center at 0 and std. dev. of "1".
Use your z-chart to determine the z-values that bracket 90% of the area
under the standard normal curve. Those values are -1.65 and +1.65.
Now find the termperatures on your 1st curve that correspond to those
z-values using the formuls z(x)=[x-mu]/sigma where z=1.65 or -1.65, mu=76 and
sigma=5.7.
1.65=[x-76]/5.7
x=(5.7)(1.65)+76
x=85.405
Also x=(5.7)(-1.65)+76
x=66.595
So the range of termperatures between 66.505 and 85.4-5 contain 90% of the
termperatures for that city.
Cheers,
Stan H.