Binomial Probability Distributions

Question

Assume that when adults with smartphones are randomly​ selected,

43​%

use them in meetings or classes. If

15

adult smartphone users are randomly​ selected, find the probability that fewer than

5

of them use their smartphones in meetings or classes

Answer 1

First, define the binomial distribution's probability mass function (PMF):

P(x) = nCx p^x q^{n-x}.

Next, correspond the given values to the variables in the problem:

p = .43 = the probability a randomly selected adult will use their phone is class/meetings

n = 15 = the number of the randomly selected adults (your sample size)

q = 1 - .43 = .57 = the probability a randomly selected adult won't use their smart phone

As for x we will not define this just yet. The questions is asking for the probability that fewer than 5 will use their smart phones, that is, either 0,1,2,3, or 4 adults. This means we want to sum over the probabilities that 0,1,2,3, or 4 adults will use their smart phones. Notice that these events are disjoint, that is, we cannot have both 0 and 4 adults who have used their smart phone over the duration of the class/meeting, which allows us to sum the probabilities.

So, what we want is:

P(0) + P(1) + P(2) + P(3) + P(4).

This is the probability that fewer than 5 adults will use their smart phones in class/meeting. To do this manually, simply plug in the values you have, with the corresponding x from {0,1,2,3,4} and sum these probabilities, e.g. P(0) = 15C0 * .43^0 * q^15, P(1) = 15C1 * .43^1 * q^14, and so on.

In doing this, you should get 0.1545517. This answer should feel somewhat intuitive, since the probability that an adult will use their smartphone is somewhat high (43%). Thus the probability that <5 will use their smartphones out of 15 should be small, since the idea is that at least 43% of them should be using their smartphones based on the probabilities we are given.

To do this in R, input:

> pbinom(4, 15, .43)