Let's solve each part of the question step-by-step:
Given the distribution of the number of devices per home:
[tex]\[
\begin{array}{|c|c|}
\hline 0 & 12 \\
\hline 1 & 19 \\
\hline 2 & 22 \\
\hline 3 & 25 \\
\hline 4 & 15 \\
\hline 5 & 7 \\
\hline \text{Total} & 100 \\
\hline
\end{array}
\][/tex]
a) The probability that a randomly selected home has 2 devices:
The probability can be found by dividing the number of homes with 2 devices by the total number of homes.
[tex]\[
\text{Probability (2 devices)} = \frac{\text{Number of homes with 2 devices}}{\text{Total number of homes}} = \frac{22}{100} = 0.22
\][/tex]
So,
[tex]\[
\boxed{0.22}
\][/tex]
b) The probability that a randomly selected home does not have 2 devices:
This probability is the complement of the probability of having 2 devices, which is 1 minus the probability of having 2 devices.
[tex]\[
\text{Probability (not 2 devices)} = 1 - \text{Probability (2 devices)} = 1 - 0.22 = 0.78
\][/tex]
So,
[tex]\[
\boxed{0.78}
\][/tex]
c) The probability that a randomly selected home has more than 2 devices:
To find this probability, we sum the probabilities of having 3, 4, and 5 devices and then divide by the total number of homes.
[tex]\[
\text{Number of homes with more than 2 devices} = 25 + 15 + 7 = 47
\][/tex]
[tex]\[
\text{Probability (more than 2 devices)} = \frac{47}{100} = 0.47
\][/tex]
So,
[tex]\[
\boxed{0.47}
\][/tex]
d) The probability that a randomly selected home has 2 or fewer devices:
To find this probability, we sum the probabilities of having 0, 1, and 2 devices and then divide by the total number of homes.
[tex]\[
\text{Number of homes with 2 or fewer devices} = 12 + 19 + 22 = 53
\][/tex]
[tex]\[
\text{Probability (2 or fewer devices)} = \frac{53}{100} = 0.53
\][/tex]
So,
[tex]\[
\boxed{0.53}
\][/tex]
To summarize:
- a) The probability that the randomly selected home has 2 devices = [tex]\(\boxed{0.22}\)[/tex]
- b) The probability that the randomly selected home does not have 2 devices = [tex]\(\boxed{0.78}\)[/tex]
- c) The probability that the randomly selected home has more than 2 devices = [tex]\(\boxed{0.47}\)[/tex]
- d) The probability that the randomly selected home has 2 or fewer devices = [tex]\(\boxed{0.53}\)[/tex]
