Who is the king of the late night TV? An internet survey estimates that, when given a choice between David Letterman and Jay Leno, 52% of the population prefers to watch Jay Leno. Suppose that you randomly select three late night TV watchers and ask them which of the two talk show hosts they prefer.

a-Find the probability distribution for x, the number of people in the sample of three who would prefer Jay Leno.
b-What is the probability that exactly one of the three would prefer Jay Leno?
c-What are the population mean and standard deviation for the random variable x?

Use a normal approximation to the binomial distribution.

Your values are the following:
p = .52, q = 1 - p = .48, x = 1, and n = 3

Find mean and standard deviation.
mean = np = (3)(.52) = ?
sd = √npq = √(3)(.52)(.48) = ?

I'll let you finish the calculations.

Once you have the mean and standard deviation, use z-scores and z-table to find probability.

P(0.5 < x < 1.5) =
P[(0.5 - mean)/sd] < z < [(1.5 - mean)/sd]

Once you have the two z-scores (substitute the mean and sd calculated), find the difference between the two using a z-table. This will be your probability.

I hope this will help get you started.

a- To find the probability distribution for x, the number of people who prefer Jay Leno in a sample of three, we can use the binomial distribution formula.

The formula for the probability mass function of a binomial distribution is:

P(x) = (nCx) * p^x * (1-p)^(n-x)

Where:
- n is the number of trials (in this case, the number of people in the sample which is 3).
- x is the number of successful outcomes (people preferring Jay Leno).
- p is the probability of a successful outcome (population prefers Jay Leno).

Given that the population prefers Jay Leno 52% of the time, p = 0.52.

Now, let's calculate the probability distribution for x:

P(0) = (3C0) * (0.52)^0 * (1-0.52)^(3-0) = 0.2304

P(1) = (3C1) * (0.52)^1 * (1-0.52)^(3-1) = 0.4512

P(2) = (3C2) * (0.52)^2 * (1-0.52)^(3-2) = 0.3456

P(3) = (3C3) * (0.52)^3 * (1-0.52)^(3-3) = 0.1728

Therefore, the probability distribution for x is:

x | P(x)
--------------
0 | 0.2304
1 | 0.4512
2 | 0.3456
3 | 0.1728

b- The probability that exactly one of the three would prefer Jay Leno (P(x=1)) is 0.4512.

c- To find the population mean (μ) and standard deviation (σ) for the random variable x, we can use the formulas:

μ = np
σ = sqrt(np(1-p))

Given n = 3 and p = 0.52:

μ = 3 * 0.52 = 1.56

σ = sqrt(3 * 0.52 * (1-0.52)) = 0.749

Therefore, the population mean (μ) is 1.56 and the standard deviation (σ) is 0.749 for the random variable x.

To find the probability distribution for x, the number of people in the sample of three who prefer Jay Leno, we can use the binomial distribution formula:

P(x) = C(n, x) * p^x * (1-p)^(n-x)

Where:
- P(x) is the probability of getting x successes (people who prefer Jay Leno) in a sample size of n=3.
- C(n, x) is the combination formula, which calculates the number of ways to choose x out of n.
- p is the probability of success, which is 0.52 or 52%.
- (1-p) is the probability of failure, which is 0.48 or 48%.

a- Calculating the probability distribution for x:
Let's substitute the values into the formula for x = 0, 1, 2, and 3:

P(0) = C(3, 0) * (0.52)^0 * (0.48)^(3-0) = 1 * 1 * (0.48)^3 = 0.1106
P(1) = C(3, 1) * (0.52)^1 * (0.48)^(3-1) = 3 * 0.52 * 0.48^2 = 0.3456
P(2) = C(3, 2) * (0.52)^2 * (0.48)^(3-2) = 3 * 0.52^2 * 0.48 = 0.4448
P(3) = C(3, 3) * (0.52)^3 * (0.48)^(3-3) = 1 * (0.52)^3 * 1 = 0.0992

The probability distribution for x is as follows:
P(0) = 0.1106
P(1) = 0.3456
P(2) = 0.4448
P(3) = 0.0992

b- The probability that exactly one of the three people would prefer Jay Leno is given by P(1), which is 0.3456 or 34.56%.

c- To find the population mean (μ) and standard deviation (σ) for the random variable x, we can use the formulas:

μ = n * p
σ = √(n * p * (1-p))

Substituting the values:
μ = 3 * 0.52 = 1.56
σ = √(3 * 0.52 * 0.48) ≈ 0.717

The population mean (μ) is 1.56 and the standard deviation (σ) is approximately 0.717.

To solve this problem, we can use the binomial probability distribution formula because there are two possible outcomes (prefer Jay Leno or prefer David Letterman) for each of the three individuals. Let's break down each part of the problem step-by-step:

a) To find the probability distribution for x, we need to calculate the probability for each value of x (0, 1, 2, 3):

For x = 0:
P(x = 0) = (number of ways to choose 0 individuals who prefer Jay Leno) * (probability of each outcome)^0 * (probability of the other outcome)^(number of trials - 0)
= C(3, 0) * (0.52)^0 * (0.48)^(3-0)
= 1 * 1 * 0.48^3
= 0.11059

For x = 1:
P(x = 1) = C(3, 1) * (0.52)^1 * (0.48)^(3-1)
= 3 * 0.52 * 0.48^2
= 0.4992

For x = 2:
P(x = 2) = C(3, 2) * (0.52)^2 * (0.48)^(3-2)
= 3 * 0.52^2 * 0.48
= 0.3456

For x = 3:
P(x = 3) = C(3, 3) * (0.52)^3 * (0.48)^(3-3)
= 1 * 0.52^3 * 0.48^0
= 0.18144

Therefore, the probability distribution for x is as follows:
P(0) = 0.11059
P(1) = 0.4992
P(2) = 0.3456
P(3) = 0.18144

b) The probability that exactly one of the three individuals would prefer Jay Leno is given by P(x = 1) = 0.4992.

c) The population mean of the random variable x can be calculated using the formula:
μ = (number of trials) * (probability of success)
μ = 3 * 0.52
μ = 1.56

The population standard deviation of the random variable x can be calculated using the formula:
σ = sqrt((number of trials) * (probability of success) * (probability of failure))
σ = sqrt(3 * 0.52 * 0.48)
σ ≈ 0.731

Therefore, the population mean for x is approximately 1.56 and the population standard deviation is approximately 0.731.