 Bobbie Comstock

2021-12-28

The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.2
(a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood? Explain.
(b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.
(c) What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood?
(d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?
(e) What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox? Jillian Edgerton

Step 1
Given:
Let,
$X=\text{Number of American adult have had chickenpox by the time they reach adulthood}.$
$p=\text{probability of American adult have had chickenpox by the time they reach adulthood, we have}$
$p=0.90$
$X\sim Binomial\left(n,p\right)$
$Y=\text{Number of American adult have not had chickenpox by the time they reach adulthood}.$
$q=\text{probability of American adult have not had chickenpox by the time they reach adulthood, we have}$
$q=1-p=0.10$
$Y\sim Binomial\left(n,q\right)$
Step 2
Solution:
(d) Let, $n=10$
$X\sim Binomial\left(n=10,p=0.9\right)$
Now,
$P\left(X\ge 1\right)=1-P\left(x=0\right)$
$=1-\left(\begin{array}{c}10\\ 0\end{array}\right)\left(0.9{\right)}^{0}\left(0.1{\right)}^{10}$
$=0.9999\approx 1$
$P\left(X\ge 1\right)=1$
(e) Let,
$Y\sim Binomial\left(n=10,q=0.1\right)$
Now,
$P\left(X\le 3\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)=P\left(x=3\right)$
$=\left(\begin{array}{c}10\\ 0\end{array}\right)\left(0.1{\right)}^{0}\left(0.9{\right)}^{10}+\left(\begin{array}{c}10\\ 1\end{array}\right)\left(0.1{\right)}^{1}\left(0.9{\right)}^{9}+\left(\begin{array}{c}10\\ 2\end{array}\right)\left(0.1{\right)}^{2}\left(0.9{\right)}^{8}+\left(\begin{array}{c}10\\ 3\end{array}\right)\left(0.1{\right)}^{3}\left(0.9{\right)}^{7}$
$P\left(X\le 3\right)=0.9872$ Serita Dewitt

Step 1
a) Yes. It fits all 4 criteria: fixed number of trials $\left(n=100\right)$S; the trials are independent (randomly selected); each trial is either a “success” (vaccinated) or a “failure” (not vaccinated); and the probability of a “success” $\left(p=0.90\right)$ is the same for all trials. Let ${C}_{100}$ be the number of people out of 100 that have had chickenpox.
b) Using the formula for binimial random variables:
$P\left({C}_{100}=97\right)=\left(\begin{array}{c}100\\ 97\end{array}\right)×{0.90}^{97}×\left(1-0.90{\right)}^{3}=0.0059$
c) Same as b.
d) Let ${C}_{10}$ be the number of people out of 10 that are vaccinated. We want:
$P\left({C}_{10}\ge 1\right)=1-P\left({C}_{10}=0\right)$
Using the binomial distribution formula:
$P\left({C}_{10}\ge 1\right)=1-P\left({C}_{10}=0\right)=1-\left(\begin{array}{c}10\\ 0\end{array}\right)×{0.90}^{0}×\left(1-0.90{\right)}^{10}=1-{10}^{-10}\approx 1$
e) We want: $P\left({C}_{10}\ge 7\right)$ Using the binomial distribution formula:
$P\left({C}_{10}\ge 7\right)=\sum _{c=7}^{10}\left(\begin{array}{c}10\\ c\end{array}\right)×{0.90}^{c}×\left(1-0.90{\right)}^{10-c}=9.12×{10}^{-6}$ karton

Step 1
Given:
p=90%=0.9, the probability that an adult has had chickenpox by age 50.
Therefore,
q=1-p=0.1, the probability that an adult has not had chickenpox by age 50.
Part (a)
Because there are only two answers: "Yes" or "No" to whether an adult has had chickenpox by age 50, the use of the binomial distribution is justified.
Part (b):
Calculate the probability that exactly 97 out of 100 sampled adults have had chickenpox.
The probability is
${P}_{1}{=}_{100}{C}_{97}\left(0.9{\right)}^{97}\left(0.1{\right)}^{3}=0.0059$
Part (c)
heis probability is equal to
${P}_{2}-{P}_{1}=1-0.006=0.994$
Part (d)
Calculate the probability that at least 1 out of 10 randomly selected adults have had chickenpox.
The probability is
${P}_{3}{=}_{10}{C}_{0}\left(0.9{\right)}^{0}\left(0.1{\right)}^{10}{+}_{10}{C}_{1}\left(0.9{\right)}^{1}\left(0.1{\right)}^{1}={10}^{-10}+{10}^{-9}={10}^{-9}\approx 0$
Part (e)
Calculate the probability that at most 3 out of 10 randomly selected adults have not had chickenpox.
The probability is
${P}_{4}=1-{\left[}_{10}{C}_{0}\left(0.9{\right)}^{0}\left(0.1{\right)}^{10}{+}_{10}{C}_{1}\left(0.9{\right)}^{1}\left(0.1{\right)}^{9}{+}_{10}{C}_{2}\left(0.9{\right)}^{2}\left(0.1{\right)}^{8}{+}_{10}{C}_{3}\left(0.9{\right)}^{3}\left(0.1{\right)}^{7}\right]$
$=1-\left({10}^{-10}+9×{10}^{-9}+3.645×{10}^{-7}+8.748×{10}^{-6}\right)$
=1 Andre BalkonE

(a) No, the use of the binomial distribution is not appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood. The binomial distribution assumes independent trials with a fixed probability of success, but in this case, the probability of having had chickenpox varies among individuals and is not constant.
(c) Assuming the probability of not having had chickenpox in childhood is 10%, we can use the binomial distribution to calculate the probability that exactly 3 out of 100 American adults have not had chickenpox in their childhood:
$P\left(X=3\right)=\left(\genfrac{}{}{0}{}{100}{3}\right)×\left(0.10{\right)}^{3}×\left(0.90{\right)}^{97}$
(d) To calculate the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox, we can use the complementary probability:
where $P\left(\text{none}\right)$ is the probability that none of the 10 adults have had chickenpox. Assuming independence, the probability that an individual has not had chickenpox is 0.10, so:
$P\left(\text{none}\right)=\left(0.10{\right)}^{10}$
(e) To calculate the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox, we can sum the probabilities of having 0, 1, 2, or 3 adults who have not had chickenpox:
where $X$ follows a binomial distribution with parameters $n=10$ and $p=0.10$. Jazz Frenia

Step 1:
(a) The use of the binomial distribution is appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood. The binomial distribution is suitable when we have a fixed number of independent trials (in this case, 100 adults) and each trial has only two possible outcomes (had chickenpox or did not have chickenpox). The probability of success (having chickenpox) remains the same for each trial, and the trials are assumed to be independent.
Step 2:
(b) To calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood, we can use the binomial probability formula. The probability mass function (PMF) for the binomial distribution is given by:
$P\left(X=k\right)=\left(\genfrac{}{}{0}{}{n}{k}\right)·{p}^{k}·\left(1-p{\right)}^{n-k}$
where:
- $P\left(X=k\right)$ is the probability of exactly k successes,
- $n$ is the total number of trials (100 adults in this case),
- $k$ is the number of successful trials (97 adults had chickenpox),
- $p$ is the probability of success in each trial (probability of having had chickenpox).
Given that 90% of Americans have had chickenpox, the probability of having had chickenpox is $p=0.9$. Substituting these values into the formula, we can calculate the probability:
$P\left(X=97\right)=\left(\genfrac{}{}{0}{}{100}{97}\right)·{0.9}^{97}·\left(1-0.9{\right)}^{100-97}$
Calculating this expression gives us the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.
Step 3:
(c) To calculate the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood, we can use the same binomial probability formula as in part (b). However, in this case, the probability of success (having had chickenpox) is different.
Given that 90% of Americans have had chickenpox, the probability of not having had chickenpox is $1-p=1-0.9=0.1$. We can calculate the probability as follows:
$P\left(X=3\right)=\left(\genfrac{}{}{0}{}{100}{3}\right)·{0.1}^{3}·\left(1-0.1{\right)}^{100-3}$
This will give us the probability that exactly 3 out of 100 randomly sampled American adults have not had chickenpox in their childhood.
Step 4:
(d) To calculate the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox, we can use the complement rule. The complement of ''at least 1'' is ''none.'' Therefore, we can calculate the probability of none of the 10 adults having had chickenpox and then subtract it from 1 to get the desired probability.
The probability of an adult not having had chickenpox is $1-p=1-0.9=0.1$. Using the binomial probability formula, we can calculate the probability of none of the 10 adults having had chickenpox:
$P\left(X=0\right)=\left(\genfrac{}{}{0}{}{10}{0}\right)·{0.1}^{0}·\left(1-0.1{\right)}^{10-0}$
Subtracting this probability from 1 gives us the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox.
Step 5:
(e) To calculate the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox, we can sum the probabilities of having 0, 1, 2, and 3 adults who have not had chickenpox.
Using the binomial probability formula, the probabilities for each case are as follows:
$P\left(X=0\right)=\left(\genfrac{}{}{0}{}{10}{0}\right)·{0.1}^{0}·\left(1-0.1{\right)}^{10-0}$
$P\left(X=1\right)=\left(\genfrac{}{}{0}{}{10}{1}\right)·{0.1}^{1}·\left(1-0.1{\right)}^{10-1}$
$P\left(X=2\right)=\left(\genfrac{}{}{0}{}{10}{2}\right)·{0.1}^{2}·\left(1-0.1{\right)}^{10-2}$
$P\left(X=3\right)=\left(\genfrac{}{}{0}{}{10}{3}\right)·{0.1}^{3}·\left(1-0.1{\right)}^{10-3}$
Summing these probabilities will give us the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox.

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