# The proportion of students who own a cell phone on college campuses across the country has increased tremendously over the past few years. It is estimated that approximately 90% of students now own a cell phone and 10% do not own a cell phone. Fifteen students are to be selected at random from a large university. Assume that the proportion of students who own a cell phone or do not own a cell phone at this university is the same as nationwide. What is the probability that all students in a simple random sample of 15 students own a cell phone (hint: find the equivalent probability of the number of students that do not own a cell phone)?

## I am not sure about that hint.

I think it is .9^15 = .206

## a study states that for a particular area, the average income per family is $35,569 and the standard deviation is $8256. what percentage of families in that area earn below $ 25,000

## To find the probability that all students in a simple random sample of 15 students own a cell phone, we need to calculate the probability that none of the students in the sample do not own a cell phone.

Given that approximately 90% of students own a cell phone and 10% do not own a cell phone, the probability that a single student does not own a cell phone is 0.10.

Since the students in the sample are selected at random, we can assume that each student's ownership of a cell phone is independent of the others.

The probability that none of the 15 students do not own a cell phone can be calculated using the binomial probability formula:

P(X = k) = (nCr) * p^k * (1-p)^(n-k)

Where:

- P(X = k) is the probability of exactly k successes (students who do not own a cell phone)

- n is the total number of trials (students in the sample)

- r is the number of combinations of k successes out of n trials

- p is the probability of a single success (a student not owning a cell phone)

In this case:

- n = 15 (students in the sample)

- k = 0 (students who do not own a cell phone)

- p = 0.10 (probability of a single student not owning a cell phone)

Plugging in these values, the calculation is as follows:

P(X = 0) = (15C0) * 0.10^0 * (1-0.10)^(15-0)

P(X = 0) = (1) * 1 * 0.9^15

P(X = 0) = 0.9^15

Using a calculator, we can determine that 0.9^15 is approximately 0.2059.

Therefore, the probability that all students in a simple random sample of 15 students own a cell phone is approximately 0.2059 or 20.59%.

## To find the probability that all students in a simple random sample of 15 students own a cell phone, we need to find the equivalent probability of the number of students who do not own a cell phone.

Given that approximately 90% of students own a cell phone, the proportion of students who do not own a cell phone is 10%.

Now, let's calculate the probability using the binomial probability formula:

P(X = k) = (nCk) * (p^k) * (1-p)^(n-k)

Where:

P(X = k) is the probability of getting k successes

n is the total number of trials (15 students)

k is the number of successes (0 students who do not own a cell phone)

p is the probability of success on a single trial (proportion of students who do not own a cell phone = 0.1)

Substituting the values into the formula, we get:

P(X = 0) = (15C0) * (0.1^0) * (1-0.1)^(15-0)

Calculating:

P(X = 0) = (1) * (1) * (0.9^15)

P(X = 0) = 0.9^15

Therefore, the probability that all students in the random sample of 15 students own a cell phone is approximately 0.205.