Question

5. (20 pts) Suppose on a given college campus 45% of the students own an iPhone, 50% an Android smartphone and 5% some other type of phone. Let X=the number of students in a simple random sample of 15 students who own an iPhone. A. What is the probability distribution of X? Note: If this is a well-known distribution it is sufficient to name the distribution and identify the value of the parameters B. Find the probability that 8 students in the sample own an iPhone. C. Find the probability that more than half of the students in the sample own an iPhone. D. Find the mean and standard deviation of the random variable X. (20 pts) A random variable X has a binomial B(100,0.2) distribution A. Explain why it is reasonable to use a normal Nu, a) distribution for X as an approximation for the distribution of X. Identify the values of u and o. B. Find the approximate probability PCX 518) using the approximation from A. Note: You must use this approximation and show all steps. No credit is given for an answer found using your calculator's built-in probability functions.

Answer 1

5)A) It is a binomial distribution.

n = 15

p = 0.45

X ~ B(15, 0.45)

B) P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X = 8) = 15C8 * (0.45)^8 * (0.55)^7 = 0.1647

C) P(X > 7.5) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

= 15C8 * (0.45)^8 * (0.55)^7 + 15C9 * (0.45)^9 * (0.55)^6 +  15C10 * (0.45)^10 * (0.55)^5 + 15C11 * (0.45)^11 * (0.55)^4 +15C12 * (0.45)^12 * (0.55)^3 +  15C13 * (0.45)^13 * (0.55)^2 + 15C14 * (0.45)^14 * (0.55)^1 + 15C15 * (0.45)^15 * (0.55)^0

= 0.3465

D) Mean = np = 15 * 0.45 = 6.75

Standard deviation = sqrt(np(1 - p))

= sqrt(15 * 0.45 * (1 - 0.45))

= 1.9268

6)A) n = 100

p = 0.2

np = 100 * 0.2 = 20

nq = 100 * (1 - 0.2) = 80

Since np > 5 and nq > 5 , so we can use normal approximation to the binomial distribution.

$\mu$ = np = 100 * 0.2 = 20

$\sigma$ = sqrt(np(1 - p))

= sqrt(100 * 0.2 * (1 - 0.2))

= 4

B) P(X < 18)

= P((X - $\mu$ )/$\sigma$< (18.5 - $\mu$ )/$\sigma$)

= P(Z < (18.5 - 20)/4)

= P(Z < -0.375)

= 0.3538