Question

NBC News reported that 1 in 20 children in the U.S. has a food allergy.† Consider selecting 15 children at random. Define the random variable x as

x = number of children in the sample of 15 that have a food allergy.

Find the following probabilities. (Round your answers to three decimal places.)

(a)

P(x < 2)

(b)

P(x ≤ 2)

(c)

P(x ≥ 3)

(d)

P(1 ≤ x ≤ 2)

Answer 1

a)

Here, n = 15, p = 0.05, (1 - p) = 0.95 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) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 * 0.95^14)
P(X < 2) = 0.463 + 0.366
P(X < 2) = 0.829

b)

Here, n = 15, p = 0.05, (1 - p) = 0.95 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) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 * 0.95^13)
P(X <= 2) = 0.463 + 0.366 + 0.135
P(X <= 2) = 0.964

c)

Here, n = 15, p = 0.05, (1 - p) = 0.95 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 3).
P(X <= 2) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 * 0.95^13)
P(X <= 2) = 0.463 + 0.366 + 0.135
P(X <= 2) = 0.964

P(X >=3) = 1 - P(x< =2)
= 1 - 0.964
= 0.036

d)

Here, n = 15, p = 0.05, (1 - p) = 0.95, x1 = 1 and x2 = 2.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(1 <= X <= 2)
P(1 <= X <= 2) = (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 * 0.95^13)
P(1 <= X <= 2) = 0.366 + 0.135
P(1 <= X <= 2) = 0.501