Question

1. (10 marks) A study of the home purchasing habits of Canadians in show homes found the following interesting facts. The probability that a person would buy a house follows a binomial distribution and that the probability that a person would buy a house was 13%. Suppose that 9 people walk into a show home that day.
a) What is the probability that one house is sold that day?

b) What is the probability that at least one house is sold that day?

c) Calculate the mean and standard deviation for the number of houses that will be bought that day?

d) Knowing that the probability of a person purchasing a house follows a binomial distribution, what could be done to increase the number of houses sold?



2. (10 marks) From past records, the number of people visiting the emergency room at a hospital follows a Poisson distribution. It is known that on average 2.2 people visit per hour.

a) What is the probability that exactly one person shows up during a 1-hour period?

b) What is the probability that at most 2 people will show up during a 1-hour period?

c) If nothing changes, what would be the mean and standard deviation of the number of people visiting the emergency room in an 8-hour time period?

Answer 1

Answer:

$1 a. P(X=1)=0.3809\\\\b. P(X\geq 1)=0.5721\\c. \mu_x=0.81, \ \ \ \sigma_x=0.7371\\d. Increase \ n$

$2.a. 0.2438\\b. 0.6227\\c. \mu_8=17.6, \ \sigma_8=4.1952$

Step-by-step explanation:

The purchase of a house has binomial distribution with p=9% and n=9

a. To find the probability of one house being sold:

$p(x)={n\choose x}p^x(1-p)^{n-x}$

The value of x is 1:

$p(x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X=x)={9\choose 1}0.09(1-0.09)^8\\\\=0.3809$

Hence, the probability  that one house is sold that day is 0.3809

b.The probability of at least one house is the same as the 1 minus the probability that no house was sold:

$p(x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 1)=1-P(X=0)\\\\=1-{9\choose 0}0.09^0(0.91)^9\\\\=1-0.4279\\\\=0.5721$

Hence,the probability that at least one house is sold that day is 0.5721

c. The mean of a binomial distribution is the product of the number of events times the probability of success=np

From a above, n=9 and p=0.09:

$\mu_x=np, \ n=9 , \ p=0.09\\\\=9\times 0.09\\\\=0.81$

#The standard deviation is calculated as np(1-p):

$\sigma_x=np(1-p),\ n=9, p=0.09\\\\=9\times 0.09\times 0.91\\\\=0.7371$

Hence, the distribution has a mean of 0.81 and a standard deviation of 0.7371

d. Given that the probability of success is low, the overall successful events can only be increased by increasing the sample size, say n should be increased to 1000.

2. a.Given a Poisson distribution with mean=2.2

-The probability in a Poisson distribution is expressed as:

$P(x;\mu)=\frac{e^{-\mu}\mu^x}{x!}$

The probability of one person showing up is calculated as:

$P(x;\mu)=\frac{e^{-\mu}\mu^x}{x!}\\\\P(1,2.2)=\frac{e^{-2.2}2.2^1}{1!}\\\\=0.2438$

Hence , the probability that exactly one person shows up during a 1-hour period is 0.2438

b. the probability that at most 2 people will show up during a 1-hour period is calculated as:

$P(x;\mu)=\frac{e^{-\mu}\mu^x}{x!}\\\\P(x\leq2 ;\mu)=P(x=0;\mu)+P(x=1;\mu)+P(x=2 ;\mu)\\\\=\frac{e^{-2.2}2.2^0}{0!}+\frac{e^{-2.2}2.2^1}{1!}+\frac{e^{-2.2}2.2^2}{2!}\\\\=0.1108+0.2438+0.2681\\\\=0.6227$

Hence,the probability that at most 2 people will show up during a 1-hour period is 0.6227

c. The mean and variance of a Poisson distribution is equal:

$E(X)=\mu=Var(X)=\sigma^2, \mu=2.2\\\\8E(X)=8(2.2)=\mu_8\\\\\mu_8=17.6\\\\V(X)=\sigma^2=17.6\\\\\sigma=4.1952$

Hence the mean after 8 hrs is 17.6 and the standard deviation will be 4.1952