The National Vaccine Information Center estimates that 90% o

Bobbie Comstock 2021-12-28 Answered
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?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Jillian Edgerton
Answered 2021-12-29 Author has 1715 answers

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?
Ask My Question
0
 
Serita Dewitt
Answered 2021-12-30 Author has 5497 answers
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:
\[P(C_{10}\geq7)=\sum_{c=7}^{10}(\begin{array}{c}10\\ c\end{array})\times0.90^{c}\times(1-0.90)^{10-c}=9.12\times10^{-6}\]
0
karton
Answered 2022-01-04 Author has 8659 answers

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
\(P_{1}=_{100}C_{97}(0.9)^{97}(0.1)^{3}=0.0059\)
Answer: 0.006 or 0.6%
Part (c)
Calculate the probability that exactly 3 adults have not had chickenpox.
heis probability is equal to
\(P_{2}-P_{1}=1-0.006=0.994\)
Answer: 0.994 or 99.4%
Part (d)
Calculate the probability that at least 1 out of 10 randomly selected adults have had chickenpox.
The probability is
\(P_{3}=_{10}C_{0}(0.9)^{0}(0.1)^{10}+_{10}C_{1}(0.9)^{1}(0.1)^{1}=10^{-10}+10^{-9}=10^{-9}\approx0\)
Answer: 0
Part (e)
Calculate the probability that at most 3 out of 10 randomly selected adults have not had chickenpox.
The probability is
\(P_{4}=1-[_{10}C_{0}(0.9)^{0}(0.1)^{10}+_{10}C_{1}(0.9)^{1}(0.1)^{9}+_{10}C_{2}(0.9)^{2}(0.1)^{8}+_{10}C_{3}(0.9)^{3}(0.1)^{7}]\)
\(=1-(10^{-10}+9\times10^{-9}+3.645\times10^{-7}+8.748\times10^{-6})\)
=1
Answer: 1.0 or 100%

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-10-29
The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.
What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood?
asked 2021-10-21
The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.
Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood.
asked 2021-10-25
The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.
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.
asked 2021-09-09
The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.
What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood?
asked 2021-09-14
The National Vaccine Information Center estimates that 90% of Americans have had chickenpox by the time they reach adulthood.
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.
asked 2021-09-26
On each of three consecutive days the National Weather Service announces that there is a 50-50 chance of rain. Assuming that the National Weather Service is correct, what is the probability that it rains on at most one of the three days? Justify your answer. (Hint: Represent the outcome that it rains on day 1 and doesn’t rain on days 2 and 3 as RNN.)
asked 2021-09-27
If a seed is planted, it has a 90% chance of growing into a healthy plant.
If 7 seeds are planted, what is the probability that exactly 2 don't grow?

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
...