Question

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 276. What is the probability that the sample proportion will be at least 2 percent more than the population proportion?

Based on historical data, your manager believes that 27% of the company's orders come from first-time customers. A random sample of 191 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.31 and 0.48?

According to a 2009 Reader's Digest article, people throw away about 9% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 67 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.14?

Answer 1

Question 1

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 276. What is the probability that the sample proportion will be at least 2 percent more than the population proportion?

Solution:

Here, we have to find P(P>0.75+0.02) = P(P>0.77)

P(P>0.77) = 1 – P(P<0.77)

Z = (P – p)/sqrt(pq/n)

We have P=0.77, p = 0.75, q = 1 – p = 1 – 0.75 = 0.25, n = 276

Z = (0.77 – 0.75)/sqrt(0.75*0.25/276)

Z = 0.02/ 0.026064

Z = 0.767342

P(Z<0.767342) = P(P<0.77) = 0.778561 (by using z-table)

P(P>0.77) = 1 – P(P<0.77)

P(P>0.77) = 1 – 0.778561

P(P>0.77) = 0.221439

Required probability = 0.221439

Question 2

Based on historical data, your manager believes that 27% of the company's orders come from first-time customers. A random sample of 191 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.31 and 0.48?

We have to find P(0.31<P<0.48)

P(0.31<P<0.48) = P(P<0.48) – P(P<0.31)

Z = (P – p)/sqrt(pq/n)

First find P(P<0.48)

We have P=0.48, p=0.27, q = 1-0.27 = 0.73, n = 191

Z = (0.48 - 0.27)/sqrt(0.27*0.73/191)

Z = 6.537213

P(Z< 6.537213) = P(P<0.48) = 1.00 (by using z-table)

Now find P(P<0.31)

Z = (0.31 - 0.27)/sqrt(0.27*0.73/191)

Z = 1.245184

P(Z<1.245184) = P(P<0.31) = 0.893468 (by using z-table)

P(0.31<P<0.48) = P(P<0.48) – P(P<0.31)

P(0.31<P<0.48) = 1.00 – 0.893468

P(0.31<P<0.48) = 0.106532

Required probability = 0.106532

Question 3

According to a 2009 Reader's Digest article, people throw away about 9% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 67 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.14?

We have to find P(P<0.14)

We are given P=0.14, p=0.09, q = 1 – 0.09 = 0.91,n=67

Z = (P – p)/sqrt(pq/n)

Z = (0.14 - 0.09) / sqrt(0.09*0.91/67)

Z = 1.430097

P(Z<1.430097) = P(P<0.14) = 0.923655 (by using z-table)

Required probability = 0.923655