Suppose that in a certain metropolitan area, 90% of all households have cable TV. Let x denote the number among four randomly selected households that have cable TV. Then x is a binomial random variable with n = 4 and p = 0.9. (Round your answers to four decimal places.)

a. Calculate p(2) = P(x = 2).

b. Interpret this probability.

i. This is the probability that at least 2 out of 4 selected households have cable TV.

ii. This is the probability that exactly 2 out of 10 selected households have cable TV.

iii. This is the probability that exactly 2 out of 4 selected households have cable TV.

iv. This is the probability that at least 2 out of 10 selected households have cable TV.

c. Calculate p(4), the probability that all four selected households have cable TV.

d. Determine P(x ≤ 3).

Step-by-step explanation:

Hello!

The variable of interest is X: Number of households that have cable tv.

This variable has a binomial distribution with parameters n=4 and p=0.90

Using the binomial distribution you have to calculate:

a. P(X=2)

I'll use the table of cumulated probabilities

P(X=2)= P(X≤2) - P(X≤1)= 0.0523 - 0.0037= 0.0486

b. There is a 4.86% probability that 2 out of 4 randomly selected households have cable TV.

Correct option is

iii. This is the probability that exactly 2 out of 4 selected households have cable TV.

c.

P(X=4)= P(X≤4)-P(X≤3)= 1 -0.3439= 0.6561

d.

Directly from the binomial table

P(x ≤ 3)= 0.3439

I hope this helps!

it would more likely be bald eagle because its has more in the box so about 75%.

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answer: plz give a pic and answer choices so i can answer!

step-by-step explanation: