Question

the centers for disease control and prevention (cdc) estimates that 11.8% of american adults suffer from chronic sinusitis (inflammation of the sinus). a random sample of 15 american are selected. round answers to at least 4 decimal places.
a) compute the probability that exactly 4 in the sample suffer from chronic sinusitis.
b) compute the probability that there are fewer than 3 in the sample that suffer from chronic sinusitis.
c) compute the probability that there are more than 3 in the sample that suffer from chronic sinusitis.
d) compute the probability that there are at most 4 in the sample that suffer from chronic sinusitis.
e) compute the mean number of americans that suffer from chronic sinusitis. 1.7700
f) compute the standard deviation of the number of americans that suffer from chronic sinusitis.

Explanation:

This is a binomial - probability problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step1: Identify the values

We have $n = 15$ (sample size), $p=0.118$ (probability of an American adult having chronic sinusitis).

Step2: Calculate part a

For $k = 4$:
First, calculate the combination $C(15,4)=\frac{15!}{4!(15 - 4)!}=\frac{15!}{4!11!}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365$.
Then, $P(X = 4)=C(15,4)\times(0.118)^{4}\times(1 - 0.118)^{15 - 4}$
$=1365\times(0.118)^{4}\times(0.882)^{11}$
$=1365\times0.000202\times0.29677$
$\approx0.0811$.

Step3: Calculate part b

$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$
$P(X = 0)=C(15,0)\times(0.118)^{0}\times(0.882)^{15}=1\times1\times0.1477\approx0.1477$
$C(15,1)=\frac{15!}{1!(15 - 1)!}=15$
$P(X = 1)=C(15,1)\times(0.118)^{1}\times(0.882)^{14}=15\times0.118\times0.1674\approx0.2971$
$C(15,2)=\frac{15!}{2!(15 - 2)!}=\frac{15\times14}{2\times1}=105$
$P(X = 2)=C(15,2)\times(0.118)^{2}\times(0.882)^{13}=105\times0.0139\times0.1898\approx0.2797$
$P(X\lt3)=0.1477 + 0.2971+0.2797=0.7245$.

Step4: Calculate part c

$P(X\gt3)=1 - P(X\leq3)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3))$
$C(15,3)=\frac{15!}{3!(15 - 3)!}=455$
$P(X = 3)=C(15,3)\times(0.118)^{3}\times(0.882)^{12}=455\times0.00166\times0.2152\approx0.1627$
$P(X\leq3)=0.1477+0.2971 + 0.2797+0.1627=0.8872$
$P(X\gt3)=1 - 0.8872 = 0.1128$.

Step5: Calculate part d

$P(X\leq4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)$
$P(X\leq4)=0.8872+0.0811=0.9683$.

Step6: Calculate part f

The mean of a binomial distribution is $\mu=np=15\times0.118 = 1.77$ (already given correctly).
The standard - deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}=\sqrt{15\times0.118\times(1 - 0.118)}=\sqrt{1.77\times0.882}=\sqrt{1.56114}\approx1.2494$.

Answer:

a) $0.0811$
b) $0.7245$
c) $0.1128$
d) $0.9683$
f) $1.2494$