Question

In: Math

When an opinion poll calls landline telephone numbers at random, approximately 30% of the numbers are...

When an opinion poll calls landline telephone numbers at random, approximately 30% of the numbers are working residential phone numbers. The remainder are either non-residential, non-working, or computer/fax numbers. You watch the random dialing machine make 20 calls. (Round your answers to four decimal places.)

(a) What is the probability that exactly 4 calls reach working residential numbers?


(b) What is the probability that at most 4 calls reach working residential numbers?


(c) What is the probability that at least 4 calls reach working residential numbers?


(d) What is the probability that fewer than 4 calls reach working residential numbers?


(e) What is the probability that more than 4 calls reach working residential numbers?

Solutions

Expert Solution

a)

Here, n = 20, p = 0.3, (1 - p) = 0.7 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 4)
P(X = 4) = 20C4 * 0.3^4 * 0.7^16
P(X = 4) = 0.1304
0

b)

Here, n = 20, p = 0.3, (1 - p) = 0.7 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 4).
P(X <= 4) = (20C0 * 0.3^0 * 0.7^20) + (20C1 * 0.3^1 * 0.7^19) + (20C2 * 0.3^2 * 0.7^18) + (20C3 * 0.3^3 * 0.7^17) + (20C4 * 0.3^4 * 0.7^16)
P(X <= 4) = 0.0008 + 0.0068 + 0.0278 + 0.0716 +
P(X <= 4) = 0.2374

c)


Here, n = 20, p = 0.3, (1 - p) = 0.7 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 4).
P(X >= 4) = 1- P(x< =3)

= 1 - ((20C0 * 0.3^0 * 0.7^20) + (20C1 * 0.3^1 * 0.7^19) + (20C2 * 0.3^2 * 0.7^18) + (20C3 * 0.3^3 * 0.7^17))

= 1- 0.1071

= 0.8929

d)

Here, n = 20, p = 0.3, (1 - p) = 0.7 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X < 4).
P(X < 4) = (20C0 * 0.3^0 * 0.7^20) + (20C1 * 0.3^1 * 0.7^19) + (20C2 * 0.3^2 * 0.7^18) + (20C3 * 0.3^3 * 0.7^17)
P(X < 4) = 0.0008 + 0.0068 + 0.0278 + 0.0716
P(X < 4) = 0.1070


e)

ere, n = 20, p = 0.3, (1 - p) = 0.7 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X > 4)

P(X> 4) = 1 - P(x< =4)
= 1 - 0.2374
= 0.7626


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