# The National Vaccine Information Center estimates that 90% o

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?

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Jillian Edgerton

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

###### Not exactly what you’re looking for?
Serita Dewitt
Step 1
a) Yes. It fits all 4 criteria: fixed number of trials $$\displaystyle{\left({n}={100}\right)}$$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” $$\displaystyle{\left({p}={0.90}\right)}$$ is the same for all trials. Let $$\displaystyle{C}_{{{100}}}$$ be the number of people out of 100 that have had chickenpox.
b) Using the formula for binimial random variables:
$P(C_{100}=97)=(\begin{array}{c}100\\ 97\end{array})\times0.90^{97}\times(1-0.90)^{3}=0.0059$
c) Same as b.
d) Let $$\displaystyle{C}_{{{10}}}$$ be the number of people out of 10 that are vaccinated. We want:
$$\displaystyle{P}{\left({C}_{{{10}}}\geq{1}\right)}={1}-{P}{\left({C}_{{{10}}}={0}\right)}$$
Using the binomial distribution formula:
$P(C_{10}\geq1)=1-P(C_{10}=0)=1-(\begin{array}{c}10\\ 0\end{array})\times0.90^{0}\times(1-0.90)^{10}=1-10^{-10}\approx1$
e) We want: $$\displaystyle{P}{\left({C}_{{{10}}}\geq{7}\right)}$$ Using the binomial distribution formula:
$P(C_{10}\geq7)=\sum_{c=7}^{10}(\begin{array}{c}10\\ c\end{array})\times0.90^{c}\times(1-0.90)^{10-c}=9.12\times10^{-6}$
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
$$P_{1}=_{100}C_{97}(0.9)^{97}(0.1)^{3}=0.0059$$
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
$$P_{3}=_{10}C_{0}(0.9)^{0}(0.1)^{10}+_{10}C_{1}(0.9)^{1}(0.1)^{1}=10^{-10}+10^{-9}=10^{-9}\approx0$$
Part (e)
Calculate the probability that at most 3 out of 10 randomly selected adults have not had chickenpox.
The probability is
$$P_{4}=1-[_{10}C_{0}(0.9)^{0}(0.1)^{10}+_{10}C_{1}(0.9)^{1}(0.1)^{9}+_{10}C_{2}(0.9)^{2}(0.1)^{8}+_{10}C_{3}(0.9)^{3}(0.1)^{7}]$$
$$=1-(10^{-10}+9\times10^{-9}+3.645\times10^{-7}+8.748\times10^{-6})$$
=1