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

Answer:

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

Step-by-step 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:

$P(x$

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

$P(x\leq3)=P(x$

b) We can calculate P(X≥4) as:

$P(X\geq4)=1-P(X$

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:

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

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

$P(x=0)=(1-p)^{n}=0.95^{50}=0.077$