Question

2. Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you.
(a) What is the probability that you need to contact four people?

(b) How many students from the college you expect to contact until you find one lives within five miles of you?

(C) What is the standard deviation of the number of students to be contacted until one says who lives within five miles of you?

(d) Suppose you randomly ask 5 students at your college, what is the probability that 3 of them live within five miles of you?

(e) Suppose you randomly ask 5 students at your college, what is the expected number of students who live within five miles of you?

How to do this five questions ? thanks​

Answer 1

Using the binomial distribution, it is found that:

a) There is a 0.0501 = 5.01% probability that you need to contact four people.

b) You expect to contact 1.82 students until you find one who lives within five miles of you.

c) The standard deviation is of 1.22 students.

d) There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

e) It is expected that 2.75 students live within five miles of you.

For each student, there are only two possible outcomes. Either they live within 5 miles of you, or they do not. The probability of a student living within 5 miles of you is independent of any other student, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• n is the number of trials.
• p is the probability of a success on a single trial.

In this problem:

• 55% of the students live within five miles of you, thus $p = 0.55$.

Item a:

This probability is P(X = 0) when n = 3(none of the first three living within five miles of you) multiplied by 0.55(the fourth does live within five miles), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$

$p = 0.091125(0.55) = 0.0501$

0.0501 = 5.01% probability that you need to contact four people.

Item b:

The expected number of trials in the binomial distribution until q successes is given by:

$E = \frac{q}{p}$

In this problem, $p = 0.55$, and 1 trial, thus $q = 1$, hence:

$E = \frac{1}{0.55} = 1.82$

You expect to contact 1.82 students until you find one who lives within five miles of you.

Item c:

The standard deviation of the number of trials until q successes are found is given by:

$S = \frac{\sqrt{q(1 - p)}}{p}$

Hence, since $q = 1, p = 0.55$:

$S = \frac{\sqrt{0.45}}{0.55} = 1.22$

The standard deviation is of 1.22 students.

Item d:

This probability is $P(X = 3)$ when $n = 5$, hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{5,3}.(0.55)^{3}.(0.45)^{2} = 0.3369$

There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

Item e:

The expected value of the binomial distribution is:

$E(X) = np$

Hence, since $n = 5, p = 0.55$:

$E(X) = 5(0.55) = 2,75$

It is expected that 2.75 students live within five miles of you.

A similar problem is given at brainly.com/question/25343741