Question

According to a Virginia Tech survey, college students make an average of 11 cell phone calls per day. Moreover, 80% of the students surveyed indicated that their parents pay their cell phone expenses (J. Elliot, “Professor Researches Cell Phone Usage Among Students,” www.physorg.com, February 26, 2007).

1. If you select a student at random, what is the probability that he or she makes more than 10 calls in a day? More than 15? More than 20?

2. If you select a random sample of 10 students, what distribution can you use to model the proportion of students who have parents who pay their cell phone expenses?

3. Using the distribution selected in (c), what is the probability that all 10 have parents who pay their cell phone expenses? At least 9? At least 8?

Answer 1

1.

Given: = 11 calls per day.

Distribution: Poisson

Formula: P(X=x) = (e^-x)/x!

The probability that he or she makes more than 10 calls in a day= P(X>10) = 1 - P(X10)

= 1 - [P(x=0)+P(x=1)+P(x=2)+......+P(x=10)] =1- 0.46 = 0.54

The probability that he or she makes more than 15 calls in a day = P(X>15) = 1 - P(X15) = 1 - 0.9074 = 0.0926

The probability that he or she makes more than 20 calls in a day = P(X>20) = 1 - P(X20) = 1 - 0.9953 = 0.0047

2.

If I select a random sample of 10 students, I can use the binomial distribution to model the proportion of students who have parents who pay their cell phone expenses.

Thus, distribution: Binomial

Formula: P(X=r) = nCr p^r q^n-r

^{p= proportion of success = 80% = 0.80}

^{q = proportion of failure = 1 - p= 0.20}

^{Sample size, n = 10}

^3.

The probability that all 10 have parents who pay their cell phone expenses = P(X=10) = 0.1074

The probability that at least 9 have parents who pay their cell phone expenses = P(X9) = P(X=9)+P(X=10) = 0.3758

The probability that at least 8 have parents who pay their cell phone expenses = P(X8) = P(X=8)+P(X=9)+P(X=10) = 0.6778