Question

Assume that when adults with smartphones are randomly selected, 59% use them in meetings or classes. If 11 adult smartphone users are randomly selected, findthe probability that fewer than 4 of them use their smartphones in meetings or classes

Answer 1

We have a population where we know that a proportion of 59% use the smartphones in meetings or classes. This can be expressed as p = 0.59.

We take a sample of n = 11 adults.

We have to calculate that 4 or fewer of adults in this sample use the smartphones in meetings or classes.

This can be modeled as a binomial model with p = 0.59 and n = 11. Then, we have to calculate this probability:

`P(k\le4)`

For binomial distributions we can express this as:

`P(k\le4)=\sum ^4_(i=0)P(x=i)=\sum ^4_(i=0)\binom{11}{i}p^i(1-p)^(11-i)`

We can calculate each term as:

`\begin{gathered} P(x=0)=\dbinom{11}{0}\cdot0.59^0\cdot0.41^(11)=1\cdot1\cdot0.0001=0.0001 \\ P(x=1)=\dbinom{11}{1}\cdot0.59^1\cdot0.41^(10)=11\cdot0.59\cdot0.0001=0.0009 \\ P(x=2)=\dbinom{11}{2}\cdot0.59^2\cdot0.41^9=55\cdot0.3481\cdot0.0003=0.0063 \\ P(x=3)=\dbinom{11}{3}\cdot0.59^3\cdot0.41^8=165\cdot0.2054\cdot0.0008=0.0271 \\ P(x=4)=\dbinom{11}{4}\cdot0.59^4\cdot0.41^7=330\cdot0.1212\cdot0.0019=0.0779 \end{gathered}`

Then, adding this probabilities, we get:

`\begin{gathered} P(x\leq4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4) \\ P(x\leq4)=0.0001+0.0009+0.0063+0.0271+0.0779 \\ P(x\leq4)=0.1121 \end{gathered}`

Then, we know that from the sample of 11 adults there is a probability P = 0.1121 that 4 or fewer adults use the smartphone for meetings or classes.

Answer: P(x≤4) = 0.1121