Question

Suppose that 92% of Sweden's population own a mobile phone.

(a) We randomly pick 12 individuals. Calculate the probability of at least 11 of them have cell phone.

(b) Now we randomly pick 1,200 individuals. Calculate approximately the probability that between 1100 and 1110 of them have a mobile phone.

Answer 1

Solution:

a) Let X be a random variable which represents the number of Sweden's individuals who own a mobile phone.

Given that 92% of Sweden's population own a mobile phone.

Hence, probability that a Sweden's citizen own a mobile phone is = 92/100 = 0.92

Let us consider "an individual from Sweden who own a mobile phone" as success. So, now we have only two mutually exclusive outcomes (success and failure).

Hence, probability of success (p) = 0.92

Number of trials (n) = 12

Since, number of trials is finite, probability of success remains constant in each trials and outcomes are independent, therefore we can consider X as binomially distributed random variable.

According to binomial probability law, the probability of exactly x successes in n trials is given by,

We have to obtain P(X = at least 11).

We have, n = 12 and p = 0.92

P(X = at least 11) = P(X ≥ 11)

P(X ≥ 11) = P(X = 11) + P(X = 12)

Using binomial probability law we get,

P(X = at least 11) = 0.7513

The probability of at least 11 of them have cell phone is 0.7513.

b) Now again, like in previous problem part we shall consider that X follows binomial distribution with parameters n = 1200 and p = 0.92.

To obtain the probability we shall use normal approximation to binomial distribution.

If np ≥ 5 and nq ≥ 5 then we can use normal approximation to binomial distribution. And then X will be approximately normally distributed with mean np and variance npq. (where, p is probability of success, q = 1 - p, and n is number of trials,).

We have, n = 1200 , p = 0.92 and q = 1 - 0.92 = 0.08

np = 1200×0.92 = 1104 which is greater than 5.

nq = 1200×0.08 = 96 which is greater than 5.

Hence, we can use normal approximation to binomial distribution to obtain the probability.

X ~ N(np, npq)

np = 1104 and npq = 1200×0.92×0.08 = 88.32

Hence, X ~ N(1104, 88.32)

We have to obtain P(1100 < X < 1110).

Also we need to apply continuity correction factor when using normal approximation to binomial distribution. The continuity correction factor is applied as follows:

P(X < n) = P(X < n - 0.5)

So now,

P(1100 < X < 1110) = P(X < 1110) - P(X < 1100)

P(1100 < X < 1110) = P(X < 1110 - 0.5) - P(X < 1100 - 0.5)

P(1100 < X < 1110) = P(X < 1109.5) - P(X < 1099.5)

Also we know that, if X~N(pq, npq) then

Using "pnorm" function of R we get,

P(Z < 0.5852) = 0.7208 and P(Z < -0.4788) = 0.3160

The probability that between 1100 and 1110 of them have a mobile phone is 0.4048.

Please rate the answer. Thank you.