Question

Binomial Distribution Exercises

Please answer all three as they are related, thank you!

4.3.4) 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 that time. Find the probability that the number of adults who have been tested for HIV in the sample would be:
a. Three
b. Less than five
c. Between five and nine, inclusive
d. More than five, but less than 10
e. Six or more

4.3.5) Refer to Exercise 4.3.4. Find the mean and variance of the number of people tested for HIV in samples of size 15.

4.3.6) Refer to Exercise 4.3.4. Suppose that we were to take a simple random sample of 25 adults today and find that two have been tested for HIV at some point in their life. Would these results be surprising? Why or why not?

NOTE: for exercise 4.3.6, you need to determine the probability of two or less of the adults had been tested for HIV

**PLEASE SHOW ALL WORK FOR ALL QUESTIONS**

Thank you!

Answer 1

X = the number of adults who have been tested for HIV has a binomial distribution

Binomial formula:
P(X) = C(n,x) * p^x * q^(n-x), where C(n,x) = n!/(x!*(n-x)!)

This is the binomial distribution

a)

P(x =3) = 15C3 * 0.32^3 *(1-0.32)^12
= 0.1457

b)
P(x<5) = P(x=0) + P(x=1) + P(x=2) +P(x=3) +P(x=4)
= 15C0* 0.32^0 *(1-0.32)^15+ 15C1 * 0.32^1 *(1-0.32)^14 + 15C2 * 0.32^2 *(1-0.32)^13 + 15C3 * 0.32^3 *(1-0.32)^12 + 15C4 * 0.32^4 *(1-0.32)^11

= 0.4477

c)

P(5 <=x< =9) = P(x<=9) - P(x < =5)
= 0.9938 - 0.6607
= 0.3331

d)

P(5 < X < 10)
= P(x=6) + P(x=7) + P(x=8) + P(x=9)
= .16706 + .10108 + .04757 + .01741
= .33312

e)

P(x> =6) = 1- P(x<=5)
= 1 - 0.6607
= 0.3393

#4.3.4

mean = np = 15 * 0.32 = 4.8

variance = np(1-p)
= 4.8 *(1-0.32) = 3.264