Quality Progress February 2005 reports on improvements in cu

Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. A key measure of customer satisfaction is the response (on a scale from 1 to 10) to the question: \"Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?\" Here, a response of 9 or 10 represents \"customer delight.\" Suppose that the survey selected 350 customers. Assume that 48% of Bank of America customers would currently express customer delight. That is, assume p = .48. (a-1) Find the probability that the sample proportion obtained from the sample of 350 Bank of America customers would be within three percentage points of the population proportion. That is, find P(.45 < Picture < .51). (Round your answer to 4 decimal places. Do not round intermediate values. Round z-value to 2 decimal places.) P(.45 < Picture < .51) (a-2) Find the probability that the sample proportion obtained from the sample of 350 Bank of America customers would be within six percentage points of the population proportion. That is, find P(.42 < Picture < .54). (Round your answer to 4 decimal places. Do not round intermediate values. Round z-value to 2 decimal places.) P(.42 < Picture < .54)

Solution

a-1.

The standard deviation of proportions is

s(p) = sqrt(p(1-p)/n) = sqrt(0.48*(1-0.48)/350) = 0.026704735

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.45      
x2 = upper bound =    0.51      
u = mean =    0.48      
          
s = standard deviation =    0.026704735      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.123396282      
z2 = upper z score = (x2 - u) / s =    1.123396282      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.130634614      
P(z < z2) =    0.869365386      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.738730772   [ANSWER]

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a-2.

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    0.42      
x2 = upper bound =    0.54      
u = mean =    0.48      
          
s = standard deviation =    0.026704735      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.246792563      
z2 = upper z score = (x2 - u) / s =    2.246792563      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.012326644      
P(z < z2) =    0.987673356      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.975346713   [ANSWER]

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