Question

NBC News reported on May 2, 2013, that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 15 children and let X be the number in the sample who have a food allergy. Then X ~ Bin(15, 0.05). (Round your probabilities to three decimal places.)

(a)

Determine both P(X ≤ 3) and P(X < 3).

P(X ≤ 3)

=

P(X < 3)

=

(b)

Determine P(X ≥ 4).

P(X ≥ 4)

=

(c)

Determine P(1 ≤ X ≤ 3).

P(1 ≤ X ≤ 3) =  

(d)

What are E(X) and σ_X? (Round your answers to two decimal places.)

E(X)

=

σ_X

=

(e)

In a sample of 70 children, what is the probability that none has a food allergy?

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

a)

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 <= 3) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 * 0.95^13) + (15C3 * 0.05^3 * 0.95^12)
P(X <= 3) = 0.4633 + 0.3658 + 0.1348 + 0.0307
P(X <= 3) = 0.995

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 < 3) = (15C0 * 0.05^0 * 0.95^15) + (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 * 0.95^13)
P(X < 3) = 0.4633 + 0.3658 + 0.1348
P(X < 3) =0.964

2)
Here, n = 15, p = 0.05, (1 - p) = 0.95 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) = (15C4 * 0.05^4 * 0.95^11) + (15C5 * 0.05^5 * 0.95^10) + (15C6 * 0.05^6 * 0.95^9) + (15C7 * 0.05^7 * 0.95^8) + (15C8 * 0.05^8 * 0.95^7) + (15C9 * 0.05^9 * 0.95^6) + (15C10 * 0.05^10 * 0.95^5)
P(X >= 4) = 0.005 + 0.001 + 0 + 0 + 0 + 0 + 0
P(X >= 4) = 0.006

c)
Here, n = 15, p = 0.05, (1 - p) = 0.95, x1 = 1 and x2 = 3.
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(1 <= X <= 3)
P(1 <= X <= 3) = (15C1 * 0.05^1 * 0.95^14) + (15C2 * 0.05^2 * 0.95^13) + (15C3 * 0.05^3 * 0.95^12)
P(1 <= X <= 3) = 0.366 + 0.135 + 0.031
P(1 <= X <= 3) = 0.532

d)

mean = np = 15 * 0.05 = 0.75

Std.dev = sqrt(npq)
= sqrt(15 * 0.05 * 0.95)
= 0.84

e)

Here, n = 70, p = 0.05, (1 - p) = 0.95 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 70C0 * 0.05^0 * 0.95^70
P(X = 0) = 0.028