SOLUTION: Suppose that the low daily temperature in February is normally districuted with a mean of 25 degrees and a standard deviation of 12 degrees. What percentage of the time will this t
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Question 1179856: Suppose that the low daily temperature in February is normally districuted with a mean of 25 degrees and a standard deviation of 12 degrees. What percentage of the time will this temperature be:
A. Below 32 degrees.
B. Below zero?
C. Between 35 and 45 degrees?
Explain. Answer by Theo(13342) (Show Source):
this is the probability that the temperature is less than 32 degrees.
the mean is 25 degrees.
the standard deviation is 12 degrees.
the z-score is (x - m) / s = (32 - 25) / 12 = 7/12 = .58333333....
area to the left of that z-score is equal to .72017.
B. Below zero?
this is the probability that the temperature is less than 0 degrees.
the mean is 25 degrees.
the standard deviation is 12 degrees.
the z-score is (x - m) / s = (0 - 25) / 12 = -25 / 12 = -2.08333333...
area to the left of that z-score is equal to .01861.
C. Between 35 and 45 degrees?
35 degree z-score is (35 - 25) / 12 = 10 / 12 = .833333333......
45 degree z-score is (45 - 25 / 12 = 20 / 12 = 1.66666666....
area to the left of .8333333..... = .79767
area to the left of 1.6666666..... = .95221
area between = larger area minus smaller area = .15454
my use of the above calculator mimics the use of the z-score tables.
the advantage is that you get an answer to more decimal places and you don't have to interpolate to get it.
i purposely only looked for the area to the left of the z-score because that's what you'd do if you were working with the tables.
here are the results of using the calculator.
it rounds the answer to 5 decimal digits.
this is more then enough detail for what is required.
for A.
for B.
for C.
note that the rounded answer for D is:
.95221 minus .79367 = .15454.
that corresponds to my more detailed answer of .1545379942.
note that, when the mean is 0 and the standard deviation is 1, the x value is the z-score.
i used z-score to mimic the work you would have had to do when using the z-score tables.
this calculator goes one better in that it can also work directly with the mean and the standard deviation.
for exmple:
for A, i entered the mean of 25 and the standard deviation of 12 and was able to get the area to the left of 32 degrees directly as shown below.
it's the same answer without you having to translate the raw score to z-score.
the calculator handles that for you.
here are the results of using the calculator directly with mean and standard deviation given rather than mean of 0 and standard deviation of 1.
if you have to use the tables rather than a calculator, i can show you how to do that.
