Question

We learned in Exercise 4.18 that about 90% of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults. (a) How many people in this sample would you expect to have had chickenpox in their childhood? And with what standard deviation? (b) Would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? How does this probability relate to your answer to part (b)?

Answer 1

Answer:

a) 108 people with a standard deviation of 3.286335

b) No

c) 0.218163 or around 21.82%

d) See explanation below.

Step-by-step explanation:

This situation can be modeled with the Binomial Distribution which gives the probability of an event that occurs exactly k times out of n, and is given by

$\large P(k;n)=\binom{n}{k}p^kq^{n-k}$

where  

$\large \binom{n}{k}$= combination of n elements taken k at a time.

p = probability that the event (“success”) occurs once

q = 1-p

In this case, the event “success” is finding an American adult who had  chickenpox before adulthood with probability p=0.9

and n=120 American adults in the sample.

The standard deviation for this binomial distribution is

$\large s=\sqrt{npq}$

where n is the sample size 120

a)

We consider a random sample of 120 American adults. How many people in this sample would you expect to have had chickenpox in their childhood?

If about 90% of American adults had chickenpox before adulthood, we expect to find 90% of 120 = 0.9*120=108 people in the sample who had chickenpox in their childhood.

The  standard deviation would be

$\large s=\sqrt{120*0.9*0.1=3.286335}$

b)

Would you be surprised if there were 105 people who have had chickenpox in their childhood?

No, because 105 and 108 are in the interval [mean - s, mean +s]

c)

What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

The probability that 105 or fewer people in this sample have had chickenpox in their childhood is

$\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}$

We compute this value easily with a spreadsheet and we get

$\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}=0.218163\approx 21.82\%$

d)

How does this probability relate to your answer to part (b)?

A binomial distribution with np>5 and nq>5 with n the sample size as is the case here, behaves pretty much like a Normal distribution with mean np and standard deviation $\large \sqrt{npq}$, so around 60% of the data are in the interval  [mean -s, mean +s] and 40% outside, so roughly 20% of the data should be in [0, mean-s]