Question

complete parts (a) and (b) below.
the number of dogs per household in a small town
dogs 0 1 2 3 4 5
probability 0.665 0.209 0.084 0.022 0.014 0.006

(a) find the mean, variance, and standard deviation of the probability distribution.
find the mean of the probability distribution.
μ = (round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for the mean of a probability distribution

The mean \(\mu\) of a discrete probability distribution is given by \(\mu=\sum_{i} x_i P(x_i)\), where \(x_i\) are the values of the random variable and \(P(x_i)\) are their corresponding probabilities.

Step2: Identify the values of \(x_i\) and \(P(x_i)\)

Here, \(x_0 = 0\), \(P(x_0)=0.665\); \(x_1 = 1\), \(P(x_1)=0.209\); \(x_2 = 2\), \(P(x_2)=0.084\); \(x_3 = 3\), \(P(x_3)=0.022\); \(x_4 = 4\), \(P(x_4)=0.014\); \(x_5 = 5\), \(P(x_5)=0.006\).

Step3: Calculate the product \(x_i P(x_i)\) for each \(i\)

• For \(x = 0\): \(0\times0.665 = 0\)
• For \(x = 1\): \(1\times0.209 = 0.209\)
• For \(x = 2\): \(2\times0.084 = 0.168\)
• For \(x = 3\): \(3\times0.022 = 0.066\)
• For \(x = 4\): \(4\times0.014 = 0.056\)
• For \(x = 5\): \(5\times0.006 = 0.03\)

Step4: Sum up all the products

\(\mu=0 + 0.209+0.168 + 0.066+0.056 + 0.03\)
\(\mu=0.209+0.168=0.377\); \(0.377 + 0.066 = 0.443\); \(0.443+0.056 = 0.499\); \(0.499 + 0.03=0.529\)
Rounding to one decimal place, \(\mu\approx0.5\) (Wait, wait, let's recalculate the sum correctly: \(0.209+0.168 = 0.377\); \(0.377+0.066=0.443\); \(0.443 + 0.056=0.499\); \(0.499+0.03 = 0.529\). Rounding to one decimal place, \(0.529\approx0.5\)? Wait, no, \(0.529\) is closer to \(0.5\) when rounded to one decimal place? Wait, \(0.529\) has the first decimal digit 5, the second is 2, so when rounding to one decimal place, we look at the second digit. Since \(2<5\), we keep the first decimal digit as 5? Wait, no: \(0.529\) is \(0.5\) (one decimal place) because the digit after 5 is 2, which is less than 5. Wait, but let's check the calculation again. Wait, maybe I made a mistake in addition. Let's add again:

\(0.209+0.168 = 0.377\)

\(0.377+0.066 = 0.443\)

\(0.443+0.056 = 0.499\)

\(0.499+0.03 = 0.529\). Yes, that's correct. So \(\mu = 0.529\approx0.5\) when rounded to one decimal place? Wait, no, \(0.529\) is \(0.5\) (one decimal) because the second decimal is 2, which is less than 5. Wait, but maybe I miscalculated the products. Wait, let's re - check the products:

• \(x = 0\): \(0\times0.665 = 0\)
• \(x = 1\): \(1\times0.209 = 0.209\)
• \(x = 2\): \(2\times0.084=0.168\) (correct)
• \(x = 3\): \(3\times0.022 = 0.066\) (correct)
• \(x = 4\): \(4\times0.014 = 0.056\) (correct)
• \(x = 5\): \(5\times0.006 = 0.03\) (correct)

Sum: \(0.209 + 0.168=0.377\); \(0.377+0.066 = 0.443\); \(0.443+0.056 = 0.499\); \(0.499+0.03=0.529\). So the mean is \(0.529\), which rounds to \(0.5\) when rounded to one decimal place.

Answer:

\(0.5\)