2. Suppose that the mean salary in a particular profession is $45,000 with a standard deviation of $1,500. What percentage of people in that profession earn less than $48,000

93%

Step-by-step explanation:

mean=45,000 standard deviation=2000 value of concern=48,000

We can easily see that since the value of concern (48,000) is GREATER than the mean, we can rule out the last two choices.

There is no possible way a number can be greater than the mean, but less than the 50th percentile.

convert 48,000 into a z-score, which is given as:

(x-mean)/standard deviation

or in this case:

(48000-45000)/2000=1.5

using my z-score table or calculator, I can see that a z-score of 1.5 corresponds to about the 93th percentile

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step-by-step explanation: