According to the Current Population Reports, published by the U.S. Bureau of the

Question

According to the Current Population Reports, published by the U.S. Bureau of the census, the mean annual alimony income received by women is $4,000. Assume a standard deviation of $7,500. Suppose 100 women are selected at random. Determine the probability that the mean annual alimony received by the 100 woman is within $500 of the population mean (sentence). Is it necessary to assume that the population of annual alimony payments is normally distributed? Explain your answer. Determine the probability that the mean annual alimony received by the 1000 woman is within $500 of the population mean (sentence). For the alimony incomes considered here, why is it necessary to take such a large sample in order to be assured of a relatively small sampling error?

Explanation / Answer

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    3500      
x2 = upper bound =    4500      
u = mean =    4000      
n = sample size =    100      
s = standard deviation =    7500      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.666666667      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.666666667      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.252492538      
P(z < z2) =    0.747507462      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.495014925   [ANSWER]  

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To get the left tailed areas of z scores, you can do in a ti-84 calculator

normcdf(-1E99, z, 0, 1)

where z is your z score.  

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