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). P(X ≤ 3) = P(X < 3) = (b) Determine P(X ≥ 4). P(X ≥ 4) = (c) Determine P(1 ≤ X ≤ 3). P(1 ≤ X ≤ 3) = (d) What are E(X) and σX? (Round your answers to two decimal places.) E(X) = σX = (e) In a sample of 90 children, what is the probability that none has a food allergy?

a) P(X ≤ 3) = 0.9946

P(X < 3) = 0.9639

b) P(X ≥ 4) = 0.0054

c) P(1 ≤ X ≤ 3) = 0.5313

d) E(X) = 0.75

σX = 0.84

e) P(X=0) = 0.0099

Step-by-step explanation:

We have x: number in the sample who have a food allergy. As the sample is of n=15 and p=0.05, we have:

a) We have to determine P(X ≤ 3) and P(X < 3)

We can calculate P(X ≤ 3) as the sum of P(0), P(1), P(2) and P(3).

P(x<3) can be calculated from the previos result as:

b) We can calculate P(X ≥ 4) as:

c) We can calculate P(1 ≤ X ≤ 3) as:

d) The expected value of a binomial variable is the product of the sample size n and the probability of success p:

The standard deviation is calculates as:

e) In this case, the sample size is n=90.

We can calculate the probability that none has a food allergy as:

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step-by-step explanation:

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