Questions: Ork: Module 2 Lesson 1 Part 1 of 3 A frequency distribution is shown below. Complete parts (a) and (b). The number of televisions per household in a small town Televisions 0 1 2 3 Households 30 447 727 1400 (a) Use the frequency distribution to construct a probability distribution. x P(x) 0 1 2 3 (Round to three decimal places as needed.)

Ork: Module 2 Lesson 1
Part 1 of 3

A frequency distribution is shown below. Complete parts (a) and (b).
The number of televisions per household in a small town
Televisions 0 1 2 3
Households 30 447 727 1400
(a) Use the frequency distribution to construct a probability distribution.
x P(x)
0 
1 
2 
3 
(Round to three decimal places as needed.)
Transcript text: Ork: Module 2 Lesson 1 Part 1 of 3 A frequency distribution is shown below. Complete parts (a) and (b). The number of televisions per household in a small town \begin{tabular}{lcccc} Televisions & 0 & 1 & 2 & 3 \\ Households & 30 & 447 & 727 & 1400 \end{tabular} (a) Use the frequency distribution to construct a probability distribution. \begin{tabular}{ll} x & $\mathrm{P}(\mathrm{x})$ \\ 0 & $\square$ \\ 1 & $\square$ \\ 2 & $\square$ \\ 3 & $\square$ \end{tabular} (Round to three decimal places as needed.)
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Solution

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Solution Steps

To construct a probability distribution from the given frequency distribution, we need to calculate the probability of each number of televisions per household. This is done by dividing the frequency of each category by the total number of households.

Step 1: Calculate the Total Number of Households

First, we calculate the total number of households by summing the frequencies of each category: \[ \text{Total households} = 30 + 447 + 727 + 1400 = 2604 \]

Step 2: Calculate the Probability Distribution

Next, we calculate the probability for each number of televisions per household by dividing the frequency of each category by the total number of households. We round each probability to four significant digits.

\[ P(0) = \frac{30}{2604} \approx 0.01152 \approx 0.012 \] \[ P(1) = \frac{447}{2604} \approx 0.1717 \approx 0.172 \] \[ P(2) = \frac{727}{2604} \approx 0.2791 \approx 0.279 \] \[ P(3) = \frac{1400}{2604} \approx 0.5375 \approx 0.538 \]

Final Answer

The probability distribution is: \[ \begin{array}{ll} x & P(x) \\ 0 & \boxed{0.012} \\ 1 & \boxed{0.172} \\ 2 & \boxed{0.279} \\ 3 & \boxed{0.538} \\ \end{array} \]

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