Q:

# Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an lg smartphone survey). if 12 adult smartphone users are randomly selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes.

Accepted Solution

A:

Explanation:

Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.

Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by

$$P(x = k) = \frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}$$

Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones).

Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by

$$P(x \ \textless \ 3) = P(x = 0) + P(x = 1) + P(x = 2) \\ \indent = \frac{12!}{0!(12-0)!}(0.54)^0 (1-0.54)^{12-0} + \frac{12!}{1!(12-1)!}(0.54)^1 (1-0.54)^{12-1} + \\ \indent \frac{12!}{2!(12-2)!}(0.54)^2 (1-0.54)^{12-2} \\ \\ \indent = \frac{12!}{(1)(12!)}(0.46)^{12} + \frac{12(11!)}{(1)(11!)}(0.54)(0.46)^{11}+ \frac{12(11)(10!)}{(2)(10!)}(0.54)^2(0.46)^{10} \\ \\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10} \\ \indent \boxed{P(x \ \textless \ 3) \approx 0.00951836732 }$$

Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.