a) To find: The mean number of new infections for the 20 vaccinated children. NS

a) Given info:
A group of 20 children are exposed to whooping cough is treated as unvaccinated. And also about 80% of the unvaccinated children exposed will develop the infections and about 5% of the vaccinated children exposed will develop the infections.
Calculation:
The random variable X represents "the number of vaccinated choldren who has been exposed to whooping will develop the infections." Here, the random sample 20 (n) children have been randomly chosen.
Also, there are two possible outcomes (vaccinated children who develop infection and vaccinated children who do not develop infection) with the probability of success "the chance that the vaccinated children who been exposed to whooping will develop the infections (p) is 0.05 and the vaccinated children who been exposed to whooping will not develop the infections is ${\displaystyle 0.95(=1-0.05)}$. Thus, X follows the binomial distribution with ${\displaystyle n=20,p=0.05}$.
Mean:
The mean is calculated by using the formula:
${\displaystyle \mu =np}$
Substitute n as 20 and p as 0.05.
${\displaystyle \mu =\left(20\right)\left(0.05\right)}$
${\displaystyle =1}$
Thus, the mean number of the new infections is 1.
The probability that no more than 2 of the 20 children develop infections:
That is, ${\displaystyle P(X\le 2)}$
Step-by-step procedure to obtain the "Binomial probability" using the MINITAB software:
- Choose Calc>Probability Distributions>Binomial Distribution.
- Choose Cumulative Probability.
- Enter Number of trials as 20 and Event probability as 0.05.
- In Input constant, enter 2.
- Click OK.
Output using the MINITAB software is given below:
Cumulative Distribution Function
Binomial with ${\displaystyle n=20\text{and}p=0.05}$
${\displaystyle xP(X\le x)}$
2 0.924516
From the output, ${\displaystyle P(X\le 2)=0.9245}$
Thus, the probability that no more than 2 of the 20 children develop infections is 0.9245.
b) Given info:
The random variable X represents "the number of vaccinated choldren who has been exposed to whooping will develop the infections." Here, the random sample 20 (n) children have been randomly chosen.
Also, there are two possible outcomes (vaccinated children who develop infection and vaccinated children who do not develop infection) with the probability of success "the chance that the vaccinated children who been exposed to whooping will develop the infections (p) is 0.80 and the vaccinated children who been exposed to whooping will not develop the infections is ${\displaystyle 0.20(=1-0.80)}$. Thus, X follows the binomial distribution with ${\displaystyle n=20,p=0.80}$.
Mean:
The mean is calculated by using the formula:
${\displaystyle \mu =np}$
Substitute n as 20 and p as 0.80.
${\displaystyle \mu =\left(20\right)\left(0.05\right)}$
${\displaystyle =16}$
Thus, the mean number of the new infections is 16.
The probability that no more than 18 of the 20 children develop infections:
That is, ${\displaystyle P(X\ge 18)=1-P\left(X<18\right)}$
${\displaystyle =1-P(X\le 17)}$
Step-by-step procedure to obtain the "Binomial probability" using the MINITAB software:
- Choose Calc>Probability Distributions>Binomial Distribution.
- Choose Cumulative Probability.
- Enter Number of trials as 20 and Event probability as 0.80.
- In Input constant, enter 17.
- Click OK.
Output using the MINITAB software is given below:
Cumulative Distribution Function
Binomial with ${\displaystyle n=20\text{and}p=0.8}$
${\displaystyle xP(X\le x)}$
17 0.793915
From the output, ${\displaystyle P(X\le 2)=0.7939}$
Therefore,
${\displaystyle P(X\ge 18)=1-P(X\le 17)}$
${\displaystyle =1-0.7939}$
${\displaystyle =0.2061}$
Thus, the probability that no more than 18 of the 20 children develop infections is 0.2061.