Question

The same survey database cited in exercise 4.3.1 (A-5) shows that 32 percent of U.S. adults indicated that they have been tested for HIV at some point in their life. Consider a simple random sample of 15 adults selected at a time. Let X be the number of adults who have been tested for HIV in the sample. For the following, find the numerical answer and describe the answer in words:
a. Three
b. Between five and nine, inclusive
c. More than five, but less than 10
d. Six or more
e. Less than five
f. Find the mean and the variance of the number of people tested for HIV in samples of size 15.

Answer 1

Answer:

a. P(X = 3) = 0.145736

b. P(5≤X≤9) = 0.5462

c. P(5<X<10) = 0.3332

d. P(X≥6) = 0.3393

e. P(X<5) = 0.4477

f. Mean = 4.8, Variance = 3.264

Step-by-step explanation:

Given - The same survey database cited in exercise 4.3.1 (A-5) shows

             that 32 percent of U.S. adults indicated that they have been

             tested for HIV at some point in their life. Consider a simple

             random sample of 15 adults selected at a time. Let X be the

             number of adults who have been tested for HIV in the sample.

To find - For the following, find the numerical answer and describe

              the answer in words:

              a. Three

              b. Between five and nine, inclusive

              c. More than five, but less than 10

              d. Six or more

              e. Less than five

              f. Find the mean and the variance of the number of people

                 tested for HIV in samples of size 15.

Proof -

Given that , n = 15, p = 32% = 0.32

Now,

a.

P(X = 3) = binomial distribution (3, 15, 0.32, 0)

             = $\left(\begin{array}{ccc}15\\3\end{array}\right)(0.32)^{3}(1-0.32)^{15 - 3}$

             = 455(0.032768)(0.009774)

             = 0.145736

⇒P(X = 3) = 0.145736

b.

P(5≤X≤9) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

               = $\left(\begin{array}{ccc}15\\5\end{array}\right)(0.32)^{5}(1-0.32)^{15 - 5}$+ $\left(\begin{array}{ccc}15\\6\end{array}\right)(0.32)^{6}(1-0.32)^{15 - 6}$ + $\left(\begin{array}{ccc}15\\7\end{array}\right)(0.32)^{7}(1-0.32)^{15 - 7}$ + $\left(\begin{array}{ccc}15\\8\end{array}\right)(0.32)^{8}(1-0.32)^{15 - 8}$+ $\left(\begin{array}{ccc}15\\9\end{array}\right)(0.32)^{9}(1-0.32)^{15 - 9}$

               = 0.213 + 0.1671 + 0.1011 + 0.0476 + 0.0174

               = 0.5462

⇒P(5≤X≤9) = 0.5462

c.

P(5<X<10) = + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

                 =  $\left(\begin{array}{ccc}15\\6\end{array}\right)(0.32)^{6}(1-0.32)^{15 - 6}$ + $\left(\begin{array}{ccc}15\\7\end{array}\right)(0.32)^{7}(1-0.32)^{15 - 7}$ + $\left(\begin{array}{ccc}15\\8\end{array}\right)(0.32)^{8}(1-0.32)^{15 - 8}$+ $\left(\begin{array}{ccc}15\\9\end{array}\right)(0.32)^{9}(1-0.32)^{15 - 9}$

               =  0.1671 + 0.1011 + 0.0476 + 0.0174

               = 0.3332

⇒P(5<X<10) = 0.3332

d.

P(X≥6) = 1 - P(X < 6)

           = 1 - P(X ≤ 5)

           = 1 - binomial(5, 15, 0.32, 1)

           = 1 - 0.6607

           = 0.3393

⇒P(X≥6) = 0.3393

e.

P(X<5) = P(X≤4)

           = binomial (4, 15, 0.32, 1)

           = 0.4477

⇒P(X<5) = 0.4477

f.

Mean = np

         = 15(0.32)

         = 4.8

⇒Mean = 4.8

Variance = np(1-p)

              = 15(0.32)(1 - 0.32)

              = 4.8(0.68)

              = 3.264

⇒Variance = 3.264