BusinessWeek surveyed students who completed their 30 studies in a master\'s pro

Question

BusinessWeek surveyed students who completed their 30 studies in a master's program (BusinessWeek, September 22, 2003). According to this survey, the average annual salary of a woman and a man 10 years after completing their studies is $ 117,000 and $ 168,000, respectively. Consider these amounts as the middle populations of wages of a woman and a man. Suppose the population standard deviation between the wages of women is $ 25,000 between the wages of men is $ 40,000. With this information: a) What is the probability that a simple random sample of 40 men the sample mean does not differ more than $ 10,000 average Population of $ 168,000? b) What is the probability that a simple random sample of 40 the sample mean women do not differ by more than $ 10,000 average Population of $ 117,000? c) Which of the two cases a) and b) are more likely to get a sample mean that differs by no more than $ 10,000 of the population mean? Why? d) What is the probability that a simple random sample of 100 Men, the sample mean does not differ by more than $ 4,000 average Population?

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 = 168000 - 10000=   158000      
x2 = upper bound = 168000 + 10000 =    178000      
u = mean =    168000      
n = sample size =    40      
s = standard deviation =    40000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.58113883      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.58113883      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.056923149      
P(z < z2) =    0.943076851      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.886153702   [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    107000      
x2 = upper bound =    127000      
u = mean =    117000      
n = sample size =    40      
s = standard deviation =    25000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.529822128      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.529822128      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.005706018      
P(z < z2) =    0.994293982      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.988587964   [ANSWER]

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c)

IT IS CASE B, FOR WOMEN.

This is so because they have less standard deviation, so they are more likely to be near the mean.  

*********************

d)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 168000 - 4000 =   164000      
x2 = upper bound = 168000 +4000 =   172000      
u = mean =    168000      
n = sample size =    100      
s = standard deviation =    40000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492   [ANSWER]  

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