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.