Answered

A recent report indicated that 22 percent of the households in a certain community speak a language other than English at home. A reporter will randomly select a household from the community until the first household that speaks a language other than English at home is selected. Let random variable Q represent the number of attempts needed until the first household that speaks a language other than English at home is selected. The random variable Q has a geometric distribution with p=0.22. Which of the following is closest to the variance of the random variable?
(A) 0.0484
(B) 2.5454
(C) 4.5455
(D) 4.0144
(E) 16.1157

Answer :

TUNDEXIGO

Answer:

(E) 16.1157

Step-by-step explanation:

Q has geometric distribution with p = 0.22

"Variance of Q" V(Q) = 1 - p / p²

V(Q) = 1 - 0.22 / 0.22²

V(Q) = 0.78 / 0.0484

V(Q) = 16.11570247933884

V(Q) = 16.1157

So, the Variance of random variable Q is 16.1157

SHRISTIPARMAR1221GO

This question is based on the variance. Hence, the correct option is D i.e. 16.1157 is closest to the variance of the random variable.

Given:

Community speak a language other than English at home  = 22 percent

Let random variable Q represent the number of attempts needed.

The random variable Q has a geometric distribution with p=0.22.

We need to determined the closest to the variance of random variable.

As we know that,

Q has a geometric distribution with p=0.22

Now calculate the variance of Q,

[tex]V(Q)=\dfrac{1-p}{p^{2} }[/tex]

[tex]V(Q)=\dfrac{1-0.22}{0.22^{2} }[/tex]

[tex]V(Q)=\dfrac{0.78}{0.484} }[/tex]

Therefore, the variance  V(Q) is 16.1157.

Hence, the correct option is D i.e. 16.1157 is closest to the variance of the random variable.

For more details, please refer this link:

https://brainly.com/question/13708253

Go Question: Other Questions