Pew Research reported that, in 2013, 78% of all teens had a cell phone. Assume this estimate is correct. We randomly pick 100 teens.

Question

What’s the approximate probability that less than three-fourths of this sample owns a cell phone?

Answer 1

In this problem, we are looking for an approximate answer, using the standard normal distribution as an approximation to the binomial. Approximation is used because the problem is stated to ask for an approximate probability, and also we can use a test with the particulars in the problem to justify.

Binomial

n = 100 trials, where the response is yes or no to having a cell phone.

p = .78, probability that a student has a cell phone.

q = 1 - p = .22

Since n*p = 78 > 5 and n*q = 22 > 5, we can justify use of the approximation.

Solution:

mean: np = 100*.78 = 78

standard deviation: square root (n * p * q) = square root (100 * .78 * .22) = 4.1425

P (x < 75)

= P ((x - mean / std dev) < (75 - 78)/ 4.4125

= P (z < -.7242)

Look up on z table, yields .2345 as the area under the normal curve to the left of the z value of -.7242)

The probability is 23.45% that fewer than 3/4 (75%) students have a cell phone.

Answer 2

Alternative calculation.

While most textbooks like to use a normal approximation for binomial distributions for ease in calculation and to normalize the lessons (no pun intended) to emphasize lessons on normal distribution probabilities, the normal approximation is just that, an approximation. An exact answer is calculable using the binomial distribution

b(x; n, P) = _nC_x * P^x * (1 - P)^{n - x} or using factorial notation

b(x; n, P) = { n! / [ x! (n - x)! ] } * P^x * (1 - P)^{n -x}

^{with ρ = 0.78, n = 100 and x = 75 the probability of any number less than 75 students out of the sample of 100 having a cell phone is calculated as 0.1972 ( which can also be written as 19.72% ) and for less than or EQUAL to 75, P = .2684.}

^{Calculation can be done by hand, using a calculator with the cumulative binomial distribution function, or an on line calculator}

Note: Since the original question asked for an approximate probability we can assume the normal approximation was the method requested as was previously posted. This exact method is presented as an extension to illustrate alternative statistical methods as well as highlighting the limitations of some commonly used estimates. (19.6% differs quite a bit from the 23.45% you get when using the normal approximation)