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. 

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.