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In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 158. What is the probability that the sample proportion will be at least 3 percent more than the population proportion?

Answer :

funmilaciousfunmie78

Answer:

The probability that the sample proportion will be at least 3 percent more than the population proportion is 0.6157

Step-by-step explanation:

We need sample proportion between 0.75 - 0.03 = 0.72 and 0.75 +0.03 = 0.78. Here we have p = 0.75 and n= 158.

So z-score for sample proportion q = 0.72

z = [tex]\frac{q - p}{\sqrt{\frac{p(1-p)}{n} } }[/tex] = [tex]\frac{0.72 - 0.75}{\sqrt{\frac{0.75(1-0.75)}{158} } }[/tex] = - [tex]\frac{0.03}{0.0344}[/tex] = - 0.872

So z-score for sample proportion q = 0.78

z= [tex]\frac{q - p}{\sqrt{\frac{p(1-p)}{n} } }[/tex] = [tex]\frac{0.78 - 0.75}{\sqrt{\frac{0.75(1-0.75)}{158} } }[/tex] =  [tex]\frac{0.03}{0.0344}[/tex] = 0.872

Therefore the probability that the sample proportion will be within 3 percent of the population proportion is

P( 0.72 < q < 0.78) = P ( -0.872 < z < 0.872)

= P( z < 0.872) - P( z < -0.872)

= 0.80785 - 0.19215

= 0.6157

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