Manhattan residents smoke in the average 11 cigarettes a day. Smoking in Manhattan has a normal distribution with standard deviation 6.27. Someone is an average smoker if he is in the middle 50% of the cigarette useage. Average smokers smoke between (how many?) and (how many?) cigs per day.
Recall that the area under the standard normal curve is always 1.00. We're interested in the area symmetric about the mean that has the area 0.50. Let's look at the total area to the left of the mean; its area is 0.50. We want to subdivide this 0.50 area so that 0.25 of this 0.50 is to the left of the mean. The question then becomes: What is the z score that corresponds to the area 0.25 measured either (1) to the left of the mean or (2) from zero to the end of the first 0.25 area.
If you happen to have a table of z-scores, look in the one for which all the z-scores are negative. What z score corresponds to an area of 0.25? Alternatively, use a calculator with built-in statistical functions. On my TI-83 Plus I have the function "invNorm( "
and if I type in "invNorm(.25) ENTER" I get z = -0.674.
This corresponds to how many cigs at the low end?
x - 11
z = -0.674 = ---------
6.27
Solve this for x: -0.674(6.27) = x - 11, or -4.226 below the mean. Then, x = 11-4.226, or 6.77. This is the low end of the middle 50% of cigarette use.
What's the high end? Use the same approach, x = 11-4.226, or 6.77, except ADD that 4.226 to the mean: x = 11+4.226, or x=15.226.
The middle 50% of cig smokers smoke between 6.77 and 15.226 cigs per day. It'd make more sense to round off these numbers: 7 to 15 per day.
