Pls answer fast pls pls pls pls

c) There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this exercise using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And $\pi$ is the probability of X happening.

In this problem, we have that:

10 students are randomly selected, so $n = 10$ .

23% of college students say they use credit cards because of the rewards program. This means that $\pi = 0.23$

(a) exactly two

This is P(X = 2).

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

There is a 29.42% probability that the number of college students who say they use credit cards because of the rewards program is exactly two.

(b) more than two

This is $P(X > 2)$ .

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities of these events must be 1. So:

$P(X \leq 2) + P(X > 2) = 1$

$P(X > 2) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

So

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

$P(X = 0) = C_{10,0}.(0.23)^{0}.(0.77)^{10} = 0.0733$

$P(X = 1) = C_{10,1}.(0.23)^{1}.(0.77)^{9} = 0.2188$

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0733 + 0.2188 + 0.2942 = 0.5863$

$P(X > 2) = 1 - P(X \leq 2) = 1 - 0.5863 = 0.4137$

There is a 41.37% probability that the number of college students who say they use credit cards because of the rewards program is more than two.

(c) between two and five inclusive.

This is

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{10,2}.(0.23)^{2}.(0.77)^{8} = 0.2942$

$P(X = 3) = C_{10,3}.(0.23)^{3}.(0.77)^{7} = 0.2343$

$P(X = 4) = C_{10,4}.(0.23)^{4}.(0.77)^{6} = 0.1225$

$P(X = 5) = C_{10,3}.(0.23)^{5}.(0.77)^{5} = 0.0439$

So

$P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2942 + 0.2343 + 0.1225 + 0.0439 = 0.6949$

There is a 69.49% probability that the number of college students who say they use credit cards because of the rewards program is between two and five, inclusive.

8 0

3 years ago

Astronomers measure large distances in light-years. One light-year is the distance that light can travel in one year, or approxi

maksim [4K]

13.6 = 10, 3, 0.6

thats the answer

6 0

2 years ago

Plz help me

ira [324]

The park charges 1.50 per ride.

1. Louisa bought 15 ride tickets at 1.50 each.

Equation: C = Entry Fee + 1.50t

t = 15.

c = 30.75

30.75 = Entry Fee + 1.50(15)

30.75 = Entry fee + 22.50

Entry fee = 30.75 - 22.50

Entry fee = $8.25

2. Entry fee was calculated as 8.25 from #1.

Tickets are 1.50 each.

He can spend 22.

22 = 8.25 + 1.50t

Subtract 8.25 from both sides:

13.75 = 1.50t

Divide both sides by 1.50:

t = 13.75 / 1.50

t = 9.2

He can buy 9 tickets.

6 0

3 years ago