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​

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

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

In this problem:

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:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125[/tex]

[tex]p = 0.091125(0.55) = 0.0501[/tex]

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:

[tex]E = \frac{q}{p}[/tex]

In this problem, [tex]p = 0.55[/tex], and 1 trial, thus [tex]q = 1[/tex], hence:

[tex]E = \frac{1}{0.55} = 1.82[/tex]

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:

[tex]S = \frac{\sqrt{q(1 - p)}}{p}[/tex]

Hence, since [tex]q = 1, p = 0.55[/tex]:

[tex]S = \frac{\sqrt{0.45}}{0.55} = 1.22[/tex]

The standard deviation is of 1.22 students.

Item d:

This probability is [tex]P(X = 3)[/tex] when [tex]n = 5[/tex], hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{5,3}.(0.55)^{3}.(0.45)^{2} = 0.3369[/tex]

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:

[tex]E(X) = np[/tex]

Hence, since [tex]n = 5, p = 0.55[/tex]:

[tex]E(X) = 5(0.55) = 2,75[/tex]

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

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

The area between z = -1.25
And z = 0.32

Happy Thanksgiving!!!!

Have a wonderful day​

Three-fourths of the difference of a number and 7 is negative fifteen. What is the number? (Please solve very clear and step by step and I'll give u brainlest) ​

SUBIECTUL I

1. 1p Să se determine primul termen al unei progresii aritmetice (an

)n≥1 știind că rația este

egală cu 5 și suma primilor zece termeni ai progresiei este egală cu 245.
2. 1p Să se determine punctele de intersecție ale graficului funcției
f: R → R, f (x) = x

2 − 7x + 12, cu axele de coordonate.
3. 1p Să se determine numărul real a pentru care vectorii
u⃗ = (a − 2)i + (3a − 1)j și v = 2i +4j sunt coliniari.
4. 1p Să se calculeze ctg x, știind că sin x =
√3
2
, x ∈ (0, 90°
).
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x
2+5
.

6. 1p Să se rezolve ecuația : Ax
2+Cx
2 = 0, x∈ N , x≥ 2.

SUBIECTUL II
1 Se consideră matricea A(a) = (

1 0 √a
1 0 1
a 1 1

), unde a ∈ (0, ∞).

a) 0.5p Arătați că det(A(4)) = 1.
b) 0.5p Demonstrați că det(A(a) ∙ A(1) − A(a + 1)) > 0, pentru orice a ∈ (0, ∞).

2. Pe mulțimea de numere reale se definește legea de compoziție asociativă
x ∘ y = √3(xy + 4) − 3(x + y).
a) 0.5p Arătați că √3 ∘ 2 = √3.
b) 0.5pDemonstrați că x ∘ y = √3(x − √3)(y − √3) + √3, pentru orice numere reale x și y.
SUBIECTUL III

1 0.5p Se consideră funcția f: R → R, f(x) = e
2x
(x − 5).

Arătați că f
′
(x) = e
2x
(2x − 9), x ∈ R .
2. 0.5p Se consideră funcția f: R → R, f(x) =
x
√x
2+1
.

Arătați că ∫ f(x)√x
2 + 1
2
0

dx = 2.

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