Question

the class is EGEN 350 pleas i need the answers of questions 7,8and 9

In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability that a person carries the gene is 0.001. a. (2pts) What is the probability that 40 people need to be tested to detect one with the gene? 7. b. (2pts) What is the expected number of people to test to detect one with the gene? (3pts) A university warehouse has received a shipment of 25 printers, of which 10 are laser printers and 15 are inkjet models. If 6 of these 25 are selected at random to be checked by a particular technician, what is the probability that exactly 2 of those selected are laser printers? 8. Heart failure is due to either natural occurrences (87%) or outside factors (13%). Natural occurrences are caused by arterial blockage, disease, and infection. Outside factors are related to induced substances or foreign objects Suppose that 20 patients will visit an emergency room with heart failure. Assume that causes of heart failure between individuals are independent. a. (3pts) What is the probability that exactly 2 individuals have conditions caused by outside factors? 9, b. (3pts) What is the probability that more than 2 individuals have conditions caused by outside factors?

Answer 1

7. The probability that a person carries the gene is p=0.001. Let X be the number of persons that need to be tested to detect one with the gene. X has a Geometric distribution with parameters, success probability (the probability that a person has the gene) p=0.001.

The probability that X=x number of persons need to be tested to detect one with the gene is (using the formula for Geometric distribution )

$\begin{align*} P(X=x)=p(1-p)^{x-1}=0.001(1-0.001)^{x-1}, \quad x=1,2,... \end{align*}$

a) The probability that 40 people need to be tested to detect one with change is

$\begin{align*} P(X=40)=0.001(1-0.001)^{40-1}=0.00096 \end{align*}$

ans: The probability that 40 people need to be tested to detect one with change is 0.0010

b) When X has a Geometric distribution the expected value of X is

$\begin{align*} E(X)=\frac{1}{p}=\frac{1}{0.001}=1000 \end{align*}$

ans: The expected number of people to test to detect one with the gene is 1000

8) there are N=25 printers, of which M=10 laser printers and N-M=25-10=15 printers are inject. A sample of K=6 printers are selected at random from this shipment of N=25 printers.

Let X be the number of laser printers in the sample of K=6 printers selected. X has a Hypergeometric distribution with parameters, N=25, M=10, K=6 and the probability that X=x printers are laser is given by

$\begin{align*} P(X=x)=\frac{\binom{M}{x}\binom{N-M}{K-x}}{\binom{N}{K}}=\frac{\binom{10}{x}\binom{25-10}{6-x}}{\binom{25}{6}}=\frac{\binom{10}{x}\binom{15}{6-x}}{\binom{25}{6}} \end{align*}$

The probability that exactly 2 of the 6 selected are laser printers is

$\begin{align*} P(X=2)&=\frac{\binom{10}{2}\binom{15}{6-2}}{\binom{25}{6}}\\ &=\frac{\frac{10!}{2!(6-2)!}\times \frac{15!}{4!(15-4)!}}{\frac{25!}{6!(25-6)!}}\\ &=0.3468 \end{align*}$

ans: The probability that exactly 2 of the 6 selected are laser printers is 0.3468

9) The probability that a randomly selected individual with heart failure, the condition is caused by outside factors is p=0.13

Let X be the number of individuals out of a total of 20 patients, that have conditions caused by outside factors.

X has a Binomial distribution with parameters, number of trials n=20, success probability (the probability that the condition is caused by outside factor) p=0.13.

The probability that X=x individuals have conditions caused by outside factor is given by (using Binomial distribution)

$\begin{align*} P(X=x)&=\binom{n}{x}p^x(1-p)^{n-x}=\binom{20}{x}0.13^x(1-0.13)^{20-x} \end{align*}$

a) The probability that exactly 2 individuals have conditions caused by outside factors is

$\begin{align*} P(X=2)&=\binom{20}{2}0.13^2(1-0.13)^{20-2}\\ &=\frac{20!}{2!(20-2)!}0.13^2(1-0.13)^{18}\\ &=0.2618 \end{align*}$

ans: The probability that exactly 2 individuals have conditions caused by outside factors is 0.2618

b) The probability that more than 2 individuals have conditions caused by outside factors is

$\begin{align*} P(X>2)&=1-P(X\le 2)\\ &=1-(P(X=0)+P(X=1)+P(X=2))\\ &=1-\left( \binom{20}{0}0.13^0(1-0.13)^{20-0}+\binom{20}{1}0.13^1(1-0.13)^{20-1}+\binom{20}{2}0.13^2(1-0.13)^{20-2}\right )\\ &=1-\left(\frac{20!}{0!(20-0)!}0.13^0(1-0.13)^{20}+\frac{20!}{1!(20-1)!}0.13^1(1-0.13)^{19}+\frac{20!}{2!(20-2)!}0.13^2(1-0.13)^{18} \right )\\ &=1-0.5080\\ &=0.4920 \end{align*}$

ans: The probability that more than 2 individuals have conditions caused by outside factors is 0.4920