Question

According to the research, 44 % of homes sold in a certain month and year were purchased by first-time buyers. A random sample of 195 people who just purchased homes is selected.

Complete parts a through e below.

a. Calculate the standard error of the proportion (Round to four decimal places as needed.)

b. What is the probability that less than 91 of them are first-time buyers? (Round to four decimal places as needed.)

c. What is the probability that more than 93 of them are 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? (Round to four decimal places as needed.)

e. What is the probability that between 74 and 80 of them are first-time buyers? (Round to four decimal places as needed.)

Answer 1

Solution:

We are given n = 195, p = 0.44

Part a

The standard error of the proportion is given as below:

Standard error = sqrt(p*q/n)

Where, q = 1 – p = 1 – 0.44 = 0.56

Standard error = sqrt(0.44*0.56/195)

Standard error = 0.035547

Part b

Here, we have to use normal approximation to binomial distribution.

Here, we have to find P(X <91) ≈ P(X<90.5)(by using continuity correction)

Mean = np = 195*0.44 = 85.8

SD = sqrt(npq) = sqrt(195*0.44*0.56) = 6.931666

Z = (X – mean) / SD

Z = (90.5 - 85.8) / 6.931666

Z = 0.678048

P(Z<0.678048) = 0.751129

(by using z-table)

Required probability = 0.7511

Part c

Here, we have to find P(X>93) ≈ P(X > 93.5) (by using continuity correction)

P(X>93.5) = 1 – P(X<93.5)

Mean = np = 195*0.44 = 85.8

SD = sqrt(npq) = sqrt(195*0.44*0.56) = 6.931666

Z = (X – mean) / SD

Z = (93.5 - 85.8) / 6.931666

Z = 1.110844

P(Z<1.110844) = P(X<93.5) = 0.866682

(by using z-table)

P(X>93.5) = 1 – P(X<93.5)

P(X>93.5) = 1 – 0.866682

P(X>93.5) = 0.133318

Required probability = 0.1333

Part d

Here, we have to find P(X>85) ≈ P(X > 85.5) (by using continuity correction)

P(X>85.5) = 1 – P(X<85.5)

Mean = np = 195*0.44 = 85.8

SD = sqrt(npq) = sqrt(195*0.44*0.56) = 6.931666

Z = (X – mean) / SD

Z = (85.5 - 85.8) / 6.931666

Z = -0.04328

P(Z<-0.04328) = P(X<85.5) = 0.482739

(by using z-table)

P(X>85.5) = 1 – P(X<85.5)

P(X>85.5) = 1 – 0.482739

P(X>85.5) = 0.517261

Required probability = 0.5173

Part e

Here, we have to find P(74<X<80)

P(74<X<80) = P(X<80) – P(X<74)

Find P(X<80)

Z = (X – mean) / SD

Z = (80 - 85.8) / 6.931666

Z = -0.83674

P(Z<-0.83674) = 0.201369

(by using z-table)

P(X<80) = 0.201369

Find P(X<74)

Z = (X – mean) / SD

Z = (74 - 85.8) / 6.931666

Z = -1.70233

P(Z<-1.70233) = P(X<74) = 0.044347

(by using z-table)

P(74<X<80) = P(X<80) – P(X<74)

P(74<X<80) = 0.201369 - 0.044347

P(74<X<80) = 0.157022

Required probability = 0.1570