A recent survey indicated that a person living in the United States consumes approximately 99 pounds of beef p

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 greater than 102 pounds?











a. 0.0379





b. 0.1181





c. 0.2323





d. 0.4296





e. 0.7140





f. 0.9908





g. none of these

A recent survey indicated that a person living in the United States consumes approximately 99 pounds of beef p
You need to use the fact that the means of samples of size n are themselves distributed with mean the same as the population from which they come and with a standard deviation of sd/(sqrt n) where sd is the standard deviation of the population. This is called the standard error of the mean.


Then you will have to use the formula (x - mean)/(se) to convert to the standard normal variable to use tables.





If you think about it really carefully which of the above answers is correct is fairly obvious as soon as you've got the standard error. You don't need tables.

teeth