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?

(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

Beginner2021-12-29Added 34 answers

Step 1

Given:

Let,

Step 2

Solution:

(d) Let,

Now,

(e) Let,

Now,

Serita Dewitt

Beginner2021-12-30Added 41 answers

Step 1

a) Yes. It fits all 4 criteria: fixed number of trials$(n=100)$ 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” $(p=0.90)$ 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:

$$P({C}_{100}=97)=(\begin{array}{c}100\\ 97\end{array})\times {0.90}^{97}\times (1-0.90{)}^{3}=0.0059$$

c) Same as b.

d) Let$C}_{10$ be the number of people out of 10 that are vaccinated. We want:

$P({C}_{10}\ge 1)=1-P({C}_{10}=0)$

Using the binomial distribution formula:

$$P({C}_{10}\ge 1)=1-P({C}_{10}=0)=1-(\begin{array}{c}10\\ 0\end{array})\times {0.90}^{0}\times (1-0.90{)}^{10}=1-{10}^{-10}\approx 1$$

e) We want:$P({C}_{10}\ge 7)$ Using the binomial distribution formula:

$$P({C}_{10}\ge 7)=\sum _{c=7}^{10}(\begin{array}{c}10\\ c\end{array})\times {0.90}^{c}\times (1-0.90{)}^{10-c}=9.12\times {10}^{-6}$$

a) Yes. It fits all 4 criteria: fixed number of trials

b) Using the formula for binimial random variables:

c) Same as b.

d) Let

Using the binomial distribution formula:

e) We want:

karton

Expert2022-01-04Added 613 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

Answer: 0.006 or 0.6%

Part (c)

Calculate the probability that exactly 3 adults have not had chickenpox.

heis probability is equal to

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

Answer: 0

Part (e)

Calculate the probability that at most 3 out of 10 randomly selected adults have not had chickenpox.

The probability is

=1

Answer: 1.0 or 100%

Andre BalkonE

Skilled2023-06-14Added 110 answers

Jazz Frenia

Skilled2023-06-14Added 106 answers

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