Question 532022
A national bank published a report on improvements in customer satisfaction and loyalty. the report concluded that 48% of the bank customers expressed customer delight regarding the banks customer services. suppose you surveyed 350 customers in your town to inquire about customer delight towards this bank.
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mean of the sample means = 0.48
std of the sample means = sqrt(0.48*0.52/350) = 0.0267
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a)find the probability that the majority of the people in your sample of 350 customers would express customer delight.:::
z(0.5) = (0.5-0.48)/0.0267) = 0.7491
P(p-hat > 0.5) = P(z > 0.7491) = 0.2269
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b)the probability of 70% that the sample proportion of respondents expressing customer delight will be less than what percentage?
invNorm(0.7) = 0.5244
p-hat = zs+u
p-hat = 0.5244*0.0267+0.48 = 0.4940
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c)the probability is 80% that the sample proportion will be contained within what symmtetrical limits of the population proportion
invNom(0.1) = -1.2816
lower limit = p-hat = -1.2816*0.0267+0.48 = 0.4458
upper limit = p-hat = +1.2816*0.0267+0.48 = 0.5142
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Cheers,
Stan H.
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