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?

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

fichoh fichoh · Answer 2 · 2021-11-07T18:53:21+01:00

Using the binomial probability principle, the solution to the posed problems given are :

0.966 ; 0.873
0.034
0.689
1.25 ; 1.090
0.0769

Given the Parameters :

P(x = x) = nCx * p^x * q^(n-x)

p = 0.05
q = 1 - 0.05 = 0.95
n = sample size = 25

1.)

P(X ≤ 3) = p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

Using a binomial probability calculator :

P(X ≤ 3) = 0.277 + 0.365 + 0.231 + 0.093 = 0.966

2.)

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

Using a binomial probability calculator

P(X < 3) = 0.277 + 0.365 + 0.231 = 0.873

3.)

P(X ≥ 4) = p(x = 4) + p(x = 5) +...+ p(x = 50)

Using a normal probability calculator

P(X ≥ 4) = 0.034

4.)

P(1 ≤ X ≤ 3) = p(x =1) + p(x =2) + p(x =3)

Using a binomial probability calculator

P(1 ≤ X ≤ 3) = 0.365 + 0.231 + 0.093 = 0.689

D.)

E(X) = np = 25 × 0.05 = 1.25

σ(X) = √(npq)

σ(X) = √(25 × 0.05 × 0.95) = √1.1875 = 1.090

E.)

Probability that none have food allergy :

P(X = 0) = 50C0 × 0.05^0 × 0.95^50

P(X = 0) = 0.0769

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