Question

According to the research,47 % of homes sold in a certain month and year were purchased by first-time buyers. A random sample of 185 people who just purchased homes is selected.

Complete parts a through e below.

a. Calculate the standard error of the proportion.

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(Round to four decimal places as needed.)

b. What is the probability that less than 89of them are first-time buyers?

P(Less than 89 of them are first-time buyers) = ?

(Round to four decimal places as needed.)

c. What is the probability that more than 95 of them are first-time buyers?

P(More than 95 of them are the first-time buyers) = ?

(Round to four decimal places as needed.)

d. What is the probability that more than 85 of them are first-time buyers?

P(More than 85of them are first-time buyers) = ?

(Round to four decimal places as needed.)

e. What is the probability that between 76 and 81 of them are first-time buyers? P(Between 76 and 81 of them are first-time buyers) = ?

(Round to four decimal places as needed.)

Answer 1

The standard error of the proportion is approximately 0.0321. The probability that less than 89 of them are first-time buyers is approximately 0.166.

Step-by-step explanation:

To calculate the standard error of the proportion, we can use the formula:

SE = √(p*(1-p)/n)

where p is the proportion of homes sold (47% or 0.47) and n is the sample size (185). Using this formula, we can calculate:

SE = √(0.47*(1-0.47)/185) ≈ 0.0321

b. To find the probability that less than 89 of the sample are first-time buyers, we can use the normal distribution. We first need to calculate the z-score, which is given by:

z = (x - np) / √(np(1-p))

where x is the number of first-time buyers (89), n is the sample size (185), and p is the proportion of first-time buyers (0.47). Plugging in these values, we find:

z = (89 - 185*0.47) / √(185*0.47*(1-0.47)) ≈ -0.9644

We can then use a standard normal distribution table or a calculator to find that the probability is approximately 0.166.