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).
b. Determine P(X >= 4). c. Determine P(1 <= X <= 3).
c. What are E(X) and Sigma (x)?
d. In a sample of 50 children, what is the probablity that none have a food allergy?

Answer 1

a) P(X<3)=0.964

P(X≤3)=0.995

b) P(X≥4)=0.005

P(1≤X≤3)=0.532

d) E(X)=0.75

Sigma=0.056

e) P(x=0)=0.077

Explanation:

We have a binomial distribution with parameters n=15 and p=0.05.

The probabiltiy that k children, out of a sample of 15 children, have a food allergy can be calculated as:

`P(x=k) = \dbinom{n}{k} p^(k)(1-p)^(n-k)\\\\\\P(x=k) = \dbinom{15}{k} 0.05^(k) 0.95^(15-k)\\\\\\`

a) P(X<3) can be calculated as the sum written as:

`P(x<3)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0) = \dbinom{15}{0} p^(0)(1-p)^(15)=1*1*0.463=0.463\\\\\\P(x=1) = \dbinom{15}{1} p^(1)(1-p)^(14)=15*0.05*0.488=0.366\\\\\\P(x=2) = \dbinom{15}{2} p^(2)(1-p)^(13)=105*0.0025*0.513=0.135\\\\\\\\P(x<3)=0.463+0.366+0.135=0.964`

We can use this result to calculate P(X≤3) as:

`P(x\leq3)=P(x<3)+P(x=3)\\\\\\P(x=3) = \dbinom{15}{3} p^(3)(1-p)^(12)=455*0.0001*0.54=0.031\\\\\\\\P(x\leq3)=0.964+0.031=0.995`

b) We can calculate P(X≥4) as:

`P(X\geq4)=1-P(X<4)=1-P(x\leq3)\\\\P(X\geq4)=1-0.995=0.005`

We can calculate P(1≤X≤3) as:

`P(1\leq x\leq3)=P(x=1)+P(x=2)+P(x=3)\\\\P(1\leq x\leq3)=0.366+0.135+0.031=0.532`

c) The mean and standard deviation can be calcalated as:

`E(x)=n\cdot p=15\cdot0.05=0.75\\\\\sigma=\sqrt{(p(1-p))/(n)}=\sqrt{(0.05\cdot0.95)/(15)}=√(0.00317)=0.056`

d) We can calculate this as the probability of a child not having an alergy, multiplied 50 times:

`P(x=0)=(1-p)^(n)=0.95^(50)=0.077`