The random variable x represents the number of cars per household in a town of 1000 households. Find the probability of randomly selecting a household that has between one and three cars, inclusive. Cars Households 135 418 2. 242 3. 122 4 83 O 0.782 O 0.218 O 0.418 O 0.135

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The random variable \( x \) represents the number of cars per household in a town of 1000 households. Find the probability of randomly selecting a household that has between one and three cars, inclusive.

| Cars | Households |
|------|------------|
| 0    | 135        |
| 1    | 418        |
| 2    | 242        |
| 3    | 122        |
| 4    | 83         |

### Options
- 0.782
- 0.218
- 0.418
- 0.135

### Explanation
The table shows the distribution of the number of cars per household. To find the probability of a household having between one and three cars, inclusive, sum the number of households with 1, 2, and 3 cars, then divide by the total number of households (1000).

\[
\text{Number of households with 1 to 3 cars} = 418 + 242 + 122 = 782
\]

The probability is calculated as follows:

\[
P(1 \leq x \leq 3) = \frac{782}{1000} = 0.782
\]
Transcribed Image Text:The random variable \( x \) represents the number of cars per household in a town of 1000 households. Find the probability of randomly selecting a household that has between one and three cars, inclusive. | Cars | Households | |------|------------| | 0 | 135 | | 1 | 418 | | 2 | 242 | | 3 | 122 | | 4 | 83 | ### Options - 0.782 - 0.218 - 0.418 - 0.135 ### Explanation The table shows the distribution of the number of cars per household. To find the probability of a household having between one and three cars, inclusive, sum the number of households with 1, 2, and 3 cars, then divide by the total number of households (1000). \[ \text{Number of households with 1 to 3 cars} = 418 + 242 + 122 = 782 \] The probability is calculated as follows: \[ P(1 \leq x \leq 3) = \frac{782}{1000} = 0.782 \]
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