A recent survey indicated that a person living in the United States consumes approximately 99 pounds of beef per year with a standard deviation of 16.9 pounds. If a sample of 100 persons living in the U.S. is selected at random, what is the probability that the average beef consumption of this sample is between 95 and 105 pounds?
a. 0.0379
b. 0.1181
c. 0.2323
d. 0.4296
e. 0.7140
f. 0.9908
g. none of these
What is a solution to this problem below?
I found (c) 0.2323 to be the answer.
Since I'm guessing you're unsure of how to solve this, use your calculator, assuming you have a TI-83, and go to 2nd VARS. Enter normalcdf(95,105,99,16.9)--you are plugging in the minimum, max, mean, and std. deviation values and you get (c). This should be correct.
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