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 25 children and let X be the number in the sample who have a food allergy. Then X ~ Bin (25,.05).

a. Determine both P (X < = 3) and P (X < 3).

b. Determine P (X > = 4). c. Determine P (1 < = X < = 3).

d. What are E (X) and Sigma (x) ?

e. In a sample of 50 children, what is the probablity that none have a food allergy?

(a) P (X<=3) = 0.9658, P (X<3) = 0.8728

(b) P (X>=4) = 0.0342

(c) P (1< = X <=3) = 0.6884

(d) E (X) = 1.25, σ (X) = 1.089

(e) P (X=0) = 0.0769

Step-by-step explanation:

We are given that X ~ Bin (25,.05) which means that this can be approximated as a binomial distribution. The formula for calculating probability using the binomial distribution is:

P (X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

x = no. of successful trials

p = probability of success

q = probability of failure = 1 - p

We have n = 25, p = 0.05 and q = 0.95

(a) P (X<=3) = P (X=0) + P (X=1) + P (X=2) + P (X=3)

= ²⁵C₀ (0.05) ⁰ (0.95) ²⁵⁻⁰ + ²⁵C₁ (0.05) ¹ (0.95) ²⁵⁻¹ + ²⁵C₂ (0.05) ² (0.95) ²⁵⁻² + ²⁵C₃ (0.05) ³ (0.95) ²⁵⁻³

= 0.2774 + 0.3649 + 0.2305 + 0.09302

P (X<=3) = 0.9658

P (X<3) = P (X=0) + P (X=1) + P (X=2)

= ²⁵C₀ (0.05) ⁰ (0.95) ²⁵⁻⁰ + ²⁵C₁ (0.05) ¹ (0.95) ²⁵⁻¹ + ²⁵C₂ (0.05) ² (0.95) ²⁵⁻²

= 0.2774 + 0.3649 + 0.2305

P (X<3) = 0.8728

(b) P (X>=4) = 1 - P (X<4)

= 1 - (P (X=0) + P (X=1) + P (X=2) + P (X=3))

= 1 - 0.9658

P (X>=4) = 0.0342

(c) P (1< = X <=3) = P (X=1) + P (X=2) + P (X=3)

= ²⁵C₁ (0.05) ¹ (0.95) ²⁵⁻¹ + ²⁵C₂ (0.05) ² (0.95) ²⁵⁻² + ²⁵C₃ (0.05) ³ (0.95) ²⁵⁻³

= 0.3649 + 0.2305 + 0.09302

P (1< = X <=3) = 0.6884

(d) E (X) = np

= (25) (0.05)

E (X) = 1.25

σ (X) = √npq

= √ (25) (0.05) (0.95)

σ (X) = 1.089

(e) n=50 and p = 0.05. We need P (X=0) so,

P (X=0) = ⁵⁰C₀ (0.05) ⁰ (0.95) ⁵⁰⁻⁰

= (1) * (1) * (0.0769)

P (X=0) = 0.0769