The National Vaccine Information Center estimates that 90% of Americans have had

chickenpox by the time they reach adulthood.

We now consider a random sample of 120 American adults:

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

b)
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 (a)?

To answer these questions, we will use the binomial distribution, which is suitable for situations where there are only two possible outcomes - success or failure. In this case, a person either had chickenpox in their childhood (success) or they didn't (failure).

a) To determine if we would be surprised by 105 people having had chickenpox in their childhood out of 120 adults, we need to assess whether this result is within the expected range.

To calculate the expected number of people who had chickenpox in their childhood, we multiply the estimated percentage by the sample size: 0.90 * 120 = 108.

The standard deviation for a binomial distribution is given by the formula: sqrt(n * p * (1 - p)), where n is the sample size and p is the estimated probability of success.

So, for our case, the standard deviation is sqrt(120 * 0.90 * (1 - 0.90)) = 3.09.

To determine the range of expected values, we can use the "68-95-99.7" rule, which states that approximately 68% of the values fall within one standard deviation of the mean, 95% fall within two standard deviations, and 99.7% fall within three standard deviations.

Within one standard deviation of the mean (108), we can expect values between 108 - 3.09 = 104.91 and 108 + 3.09 = 111.09, rounded to the nearest whole number.

Since 105 falls within this range, it is not surprising to have 105 people who had chickenpox in their childhood out of 120 adults.

b) To determine the probability that 105 or fewer people in this sample have had chickenpox in their childhood, we must sum up the probabilities of having 0, 1, 2, ..., 105 people who had chickenpox in their childhood.

This can be calculated using the binomial probability formula: P(X=k) = (n C k) * (p^k) * ((1-p)^(n-k)), where X is the random variable representing the number of successes, k is the specific number of successes we're interested in, n is the sample size, p is the estimated probability of success, and (n C k) represents the number of ways to choose k successes from n trials.

Calculating this probability for k=0, k=1, ..., k=105, we can add up these individual probabilities to get the desired probability.

However, these calculations can be time-consuming. Alternatively, we can use statistical software or online calculators that provide the binomial probability function to find the cumulative probability for this range directly.

The probability we're interested in is P(X ≤ 105), which tells us the probability of having 105 or fewer people who had chickenpox in their childhood out of the sample of 120.

By comparing this probability to our answer in part (a), if the probability is high, it means that having 105 people with chickenpox in their childhood would not be considered surprising. If the probability is low, it would be considered surprising.