Question

It is known that 15% of the population is afraid of being alone at night. If a random sample of 20 Americans is selected, what is the probability that :

a. Exactly 6 of them are afraid?

b. Less than 2 of them are afraid?

c. At least 1 of them is afraid?

d. What is the standard deviation for the number of people that are afraid?

Answer #1

a)

Here, n = 20, p = 0.15, (1 - p) = 0.85 and x = 6

As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)

We need to calculate P(X = 6)

P(X = 6) = 20C6 * 0.15^6 * 0.85^14

P(X = 6) = 0.0454

b)

Here, n = 20, p = 0.15, (1 - p) = 0.85 and x = 2

As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)

We need to calculate P(X <= 2).

P(X <2) = (20C0 * 0.15^0 * 0.85^20) + (20C1 * 0.15^1 *
0.85^19)

P(X <2) = 0.0388 + 0.1368

P(X < 2) = 0.1756

c)

Here, n = 20, p = 0.15, (1 - p) = 0.85 and x = 1

As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)

We need to calculate P(X >= 1).

We need to calculate P(X <= 0).

P(X <= 0) = (20C0 * 0.15^0 * 0.85^20)

P(X <= 0) = 0.0388

P(X <= 0) = 0.0388

P(X>=1) = 1- P(x< =0)

=1 - 0.0388

= 0.9612

d)

std.dev = sqrt(npq)

=sqrt(20 * 0.15 * 0.85)

= 1.5969

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