Use the probability distribution to complete parts​ (a) through​ (d) below.

The probability distribution of number of televisions per household in a small town

x

0

1

2

3
 
​P(x)

0.03

0.11

0.31

0.55


Let's address each part step by step using the provided probability distribution:

### Given:
- \( x \) is the number of televisions per household.
- \( P(x) \) is the probability of having \( x \) televisions.

| \( x \) | \( P(x) \) |
|:------:|:---------:|
|   0    |   0.03    |
|   1    |   0.11    |
|   2    |   0.31    |
|   3    |   0.55    |

### Part (a): Verify that the probability distribution is valid.

A probability distribution is valid if the sum of all probabilities equals 1. Let's check:

\[
\text{Sum of } P(x) = P(0) + P(1) + P(2) + P(3) = 0.03 + 0.11 + 0.31 + 0.55
\]

\[
\text{Sum of } P(x) = 1.00
\]

Since the sum is 1, the probability distribution is valid.

### Part (b): Find the mean \( \mu \) of the distribution.

The mean (expected value) \( \mu \) of a probability distribution is given by:

\[
\mu = \sum x \cdot P(x)
\]

Let's calculate:

\[
\mu = (0 \times 0.03) + (1 \times 0.11) + (2 \times 0.31) + (3 \times 0.55)
\]

\[
\mu = 0 + 0.11 + 0.62 + 1.65 = 2.38
\]

So, the mean \( \mu \) is 2.38 televisions per household.

### Part (c): Find the variance \( \sigma^2 \) and standard deviation \( \sigma \).

The variance \( \sigma^2 \) is given by:

\[
\sigma^2 = \sum \left(x - \mu\right)^2 \cdot P(x)
\]

First, calculate each term \( \left(x - \mu\right)^2 \):

\[
\sigma^2 = (0 - 2.38)^2 \times 0.03 + (1 - 2.38)^2 \times 0.11 + (2 - 2.38)^2 \times 0.31 + (3 - 2.38)^2 \times 0.55
\]

Calculating each:

\[
\sigma^2 = (5.6644 \times 0.03) + (1.9044 \times 0.11) + (0.1444 \times 0.31) + (0.3844 \times 0.55)
\]

\[
\sigma^2 = 0.169932 + 0.209484 + 0.044764 + 0.21142 = 0.6356
\]

Thus, the variance \( \sigma^2 \) is approximately 0.6356.

The standard deviation \( \sigma \) is the square root of the variance:

\[
\sigma = \sqrt{0.6356} \approx 0.7973
\]

### Part (d): Interpret the results.

- **Mean \( \mu = 2.38 \):** On average, a household in this small town has about 2.38 televisions.
- **Standard Deviation \( \sigma \approx 0.7973 \):** The number of televisions per household typically deviates from the mean by about 0.7973 televisions.

Would you like any further explanation or have more questions?

### 5 Related Questions:
1. How would the mean change if the distribution shifted towards more households having 3 televisions?
2. What is the probability that a randomly selected household has more than 1 television?
3. How would you calculate the cumulative distribution function (CDF) for this distribution?
4. What does a low standard deviation indicate about the distribution of televisions per household?
5. How would the variance change if a significant number of households owned no televisions?

**Tip:** The mean gives a central value, while the standard deviation tells us how spread out the values are from the mean. Understanding both provides a clearer picture of the data distribution.

Explore how probability distributions are applied to analyze the number of televisions per household in a small town. Learn how to verify a probability distribution, calculate the mean, variance, and standard deviation, and interpret the results. Suitable for advanced high school students studying probability theory.

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