THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Let the random variable Z follow a standard normal distribution.
-The distribution of annual incomes of a sample of college graduates is normally distributed with a mean of $52,000 and a standard deviation of 1,000.About 68 percent of the incomes lie between what two income levels?
A)52,000 and 62,000
B)50,000 and 60,000
C)51,000 and 53,000
D)53,000 and 57,000

Question

Quizplus · Accepted Answer

Since the distribution of the annual incomes is normal with a mean of $52,000 and a standard deviation of $1,000, we can convert the distribution to a standard normal distribution using the formula:
Z = (X - μ) / σ 
where X is the random variable (annual income in this case), μ is the mean and σ is the standard deviation.

Substituting the values, we get:
Z = (X - 52,000) / 1,000

Now we need to find the Z-scores corresponding to the 16th and 84th percentiles of the standard normal distribution. Using a standard normal distribution table or calculator, we find that the Z-score corresponding to the 16th percentile is -1 and the Z-score corresponding to the 84th percentile is +1.

So, we can write:
-1 = (X - 52,000) / 1,000 
Solving for X, we get X = $51,000

and

+1 = (X - 52,000) / 1,000 
Solving for X, we get X = $53,000

Therefore, about 68 percent of the incomes lie between $51,000 and $53,000.