Question

It is assumed in this section that the probability that a respondent agrees to answer the question is equal at 0.6.

1) The polling institute interviews 700 people. We denote by X the random variable corresponding to the number of people interviewees who agree to answer the question.

a) What is the law of the random variable X? Justify the answer.

b) What is the best approximation of P (X> 400) among the following numbers?

c)At least how many people must the institute interview to guarantee, with a probability greater than 0.9, the number of people responding to the survey be greater than or equal to 400.

0,92      0,93       0,94     0,95.

Answer 1

Solution

a)

X ~ (N,p)

X ~ (700, 0.6)

=> random variable X will follow a b inomial distrbution where n = 700 and p = 0.60

E[x] = np

700 * 0.60 = 420

and SD[x] = sqrt(n*p*(1-p))

sqrt(700 * 0.60 * 0.40) = 12.962

b)

P(x > 400)

z =(xbar-mean)/SD

(400 - 420)/12.9615 = -1.543

From z-table we find the p-value as

0.93859 ~ 0.94

c)

From z-table we find the inverse z score for 0.9 to be = -1.282

Let the number be N

N*0.6 = E(x)

SD = sqrt(N*0.6*0.4)

-1.282 = (400-0.6*N)/SQRT(N*0.24)

(-1.282)^2*(0.24*N) = 160000+0.36N^2-480N

0.394446N = 160000+0.36N^2 -480N

Solving for the equation we get N = 694.247~ 695 people