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).

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?

Answer 1

a.
`P(x\le 3) = 0.9957` and
`P(x < 3) = 0.965`

b.
`P(X \ge 4) = 0.0043`

c.
`P(1 \le x \le 3)= 0.5313`

d.
`E(x) = 0.75` and
`\sigma(x)= 0.8441`

e.
`P(0) = 0.0769`

Explanation:

Given

X ~ Bin(15, 0.05)

This implies that:

`n = 15`

`p =0.05`

A binomial distribution is represented as:

`P(x) = ^nC_x\ p^x\ (1- p)^(n-x)`

Solving (a):

(i)

`P(x\le 3)`

This is solved as:

`P(x\le 3) = P(0) + P(1) + P(2) + P(3)`

`P(0) = ^(15)C_0\ * 0.05^0\ (1- 0.05)^(15-0) = 1 * 1 * 0.95^(15) = 0.4633`

`P(1) = ^(15)C_1\ * 0.05^1\ (1- 0.05)^(15-1) = 15 * 0.05 * 0.95^(14) = 0.3658`

`P(2) = ^(15)C_2\ * 0.05^2\ (1- 0.05)^(15-2) = 105 * 0.05^2 * 0.95^(13) = 0.1348`

`P(3) = ^(15)C_3\ * 0.05^3\ (1- 0.05)^(15-3) = 455 * 0.05^3 * 0.95^(12) = 0.0307`

So:

`P(x\le 3) = 0.4644 + 0.3658 + 0.1348 + 0.0307`

`P(x\le 3) = 0.9957`

(ii)

`P(x < 3) = P(0) + P(1) + P(2)`

`P(x < 3) = 0.4644 + 0.3658 + 0.1348`

`P(x < 3) = 0.965`

Solving (b):

`P(X \ge 4)`

This is represented as:

`P(x \le 3) + P(X \ge 4) = 1`

`P(X \ge 4) = 1 - P(x \le 3)`

`P(X \ge 4) = 1 - 0.9957`

`P(X \ge 4) = 0.0043`

Solving (c):

`P(1 \le x \le 3)`

This is represented as:

`P(1 \le x \le 3) =P(1) + P(2) + P(3)`

`P(1 \le x \le 3) = P(1) + P(2) + P(3)`

`P(1 \le x \le 3)= 0.3658 + 0.1348 + 0.0307`

`P(1 \le x \le 3)= 0.5313`

Solving (d): E(x) and Sigma(x)

`E(x) = n * p`

`E(x) = 15 * 0.05`

`E(x) = 0.75`

`\sigma(x)= √(E(x)* (1-p))`

`\sigma(x)= √(0.75* (1 - 0.05))`

`\sigma(x)= √(0.7125)`

`\sigma(x)= 0.8441`

Solving (e):

Calculate P(0) when n= 50

Using:
`P(x) = ^nC_x\ p^x\ (1- p)^(n-x)`

`P(0) = ^(50)C_0 * 0.05^0 * (1- 0.05)^(50-0)`

`P(0) = 1 * 1 * (0.95)^(50)`

`P(0) = 1 * 1 * 0.0769`

`P(0) = 0.0769`