Sure, let's go through the solution step-by-step. Given the probability distribution of the number of cars per household in a town of 1000 households:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Cars} & \text{Households} \\
\hline
0 & 125 \\
\hline
1 & 428 \\
\hline
2 & 256 \\
\hline
3 & 108 \\
\hline
4 & 83 \\
\hline
\end{array}
\][/tex]
a) To find the probability of randomly selecting a household that has less than two cars:
1. Identify the number of households with less than two cars. This includes households with 0 cars and 1 car.
2. Sum these households: [tex]\( 125 + 428 = 553 \)[/tex]
3. Calculate the probability by dividing by the total number of households (1000):
[tex]\[
P(x < 2) = \frac{125 + 428}{1000} = \frac{553}{1000} = 0.553
\][/tex]
Therefore, [tex]\( P(x < 2) = 0.553 \)[/tex].
b) To find the probability of randomly selecting a household that has at least one car:
1. Identify the number of households with at least one car. This includes households with 1, 2, 3, or 4 cars.
2. Sum these households: [tex]\( 428 + 256 + 108 + 83 = 875 \)[/tex]
3. Calculate the probability by dividing by the total number of households (1000):
[tex]\[
P(x \geq 1) = \frac{428 + 256 + 108 + 83}{1000} = \frac{875}{1000} = 0.875
\][/tex]
Therefore, [tex]\( P(x \geq 1) = 0.875 \)[/tex].
c) To find the probability of randomly selecting a household that has between one and three cars, inclusive:
1. Identify the number of households with between one and three cars. This includes households with 1, 2, or 3 cars.
2. Sum these households: [tex]\( 428 + 256 + 108 = 792 \)[/tex]
3. Calculate the probability by dividing by the total number of households (1000):
[tex]\[
P(1 \leq x \leq 3) = \frac{428 + 256 + 108}{1000} = \frac{792}{1000} = 0.792
\][/tex]
Therefore, [tex]\( P(1 \leq x \leq 3) = 0.792 \)[/tex].
So the results are:
a) [tex]\( P(x < 2) = 0.553 \)[/tex]
b) [tex]\( P(x \geq 1) = 0.875 \)[/tex]
c) [tex]\( P(1 \leq x \leq 3) = 0.792 \)[/tex]
