the centers for disease control and prevention (cdc) estimates that 11.5% of american adults suffer from…

the centers for disease control and prevention (cdc) estimates that 11.5% of american adults suffer from chronic sinusitis (inflammation of the sinus). a random sample of 17 american are selected. round answers to at least 4 decimal places.\na) compute the probability that exactly 3 in the sample suffer from chronic sinusitis.\nb) compute the probability that there are fewer than 2 in the sample that suffer from chronic sinusitis.\nc) compute the probability that there are more than 4 in the sample that suffer from chronic sinusitis.\nd) compute the probability that there are at most 3 in the sample that suffer from chronic sinusitis.\ne) compute the mean number of americans that suffer from chronic sinusitis.\nf) compute the standard deviation of the number of americans that suffer from chronic sinusitis.

the centers for disease control and prevention (cdc) estimates that 11.5% of american adults suffer from chronic sinusitis (inflammation of the sinus). a random sample of 17 american are selected. round answers to at least 4 decimal places.\na) compute the probability that exactly 3 in the sample suffer from chronic sinusitis.\nb) compute the probability that there are fewer than 2 in the sample that suffer from chronic sinusitis.\nc) compute the probability that there are more than 4 in the sample that suffer from chronic sinusitis.\nd) compute the probability that there are at most 3 in the sample that suffer from chronic sinusitis.\ne) compute the mean number of americans that suffer from chronic sinusitis.\nf) compute the standard deviation of the number of americans that suffer from chronic sinusitis.

Answer

Explanation:

Step1: Identify the binomial - distribution parameters

Let (n = 17) (sample size), (p=0.115) (probability of success, i.e., an American adult has chronic sinusitis), and (q = 1 - p=1 - 0.115 = 0.885). The binomial probability formula is (P(X = k)=C(n,k)\times p^{k}\times q^{n - k}), where (C(n,k)=\frac{n!}{k!(n - k)!}).

Step2: Calculate part a

For (k = 3): [ \begin{align*} C(17,3)&=\frac{17!}{3!(17 - 3)!}=\frac{17\times16\times15}{3\times2\times1}=680\ P(X = 3)&=C(17,3)\times(0.115)^{3}\times(0.885)^{14}\ &=680\times0.001520875\times0.187777\ &\approx0.1937 \end{align*} ]

Step3: Calculate part b

(P(X\lt2)=P(X = 0)+P(X = 1)) [ \begin{align*} C(17,0)&=\frac{17!}{0!(17 - 0)!}=1\ P(X = 0)&=C(17,0)\times(0.115)^{0}\times(0.885)^{17}=1\times1\times0.1307=0.1307\ C(17,1)&=\frac{17!}{1!(17 - 1)!}=17\ P(X = 1)&=C(17,1)\times(0.115)^{1}\times(0.885)^{16}=17\times0.115\times0.1477\ &=17\times0.0170855\approx0.2904\ P(X\lt2)&=0.1307 + 0.2904=0.4211 \end{align*} ]

Step4: Calculate part c

(P(X\gt4)=1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))) [ \begin{align*} C(17,2)&=\frac{17!}{2!(17 - 2)!}=136\ P(X = 2)&=C(17,2)\times(0.115)^{2}\times(0.885)^{15}=136\times0.013225\times0.167\ &\approx0.2993\ C(17,4)&=\frac{17!}{4!(17 - 4)!}=2380\ P(X = 4)&=C(17,4)\times(0.115)^{4}\times(0.885)^{13}=2380\times0.00017493\times0.2122\ &\approx0.0899\ P(X\leq4)&=0.1307+0.2904 + 0.2993+0.1937+0.0899=0.904\ P(X\gt4)&=1 - 0.904 = 0.0960 \end{align*} ]

Step5: Calculate part d

(P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)) (P(X\leq3)=0.1307+0.2904 + 0.2993+0.1937=0.9141)

Step6: Calculate part e

The mean of a binomial distribution is (\mu=np). So (\mu=17\times0.115 = 1.9550)

Step7: Calculate part f

The standard - deviation of a binomial distribution is (\sigma=\sqrt{npq}). [ \begin{align*} \sigma&=\sqrt{17\times0.115\times0.885}\ &=\sqrt{1.955\times0.885}\ &=\sqrt{1.730175}\approx1.3154 \end{align*} ]

Answer:

a) (0.1937) b) (0.4211) c) (0.0960) d) (0.9141) e) (1.9550) f) (1.3154)