Complete parts (a) and (b) below.

The number of dogs per household in a small town

Dogs

0

1

2.

3

Probability 0.650 0.216 0.087

0.026

4

5

0.014

0.009

(a) Find the mean, variance, and standard deviation of the probability distribution

Find the mean of the probability distribution

(Round to one decimal place as needed.)

Answer:

a) The mean of the probability distribution = 0.6

b) The variance of the probability distribution = 0.3

c) The standard deviation of the probability distribution = 0.5

Step-by-step explanation:

a) Data and Calculations:

number of dogs per household in a small town

Dogs   Probability  Expected  Mean Diff    Mean Diff.

                               Value                            Squared

0          0.650           0               -0.569         0.32376

1           0.216            0.216        -0.353          0.12461

2         0.087            0.174         -0.395          0.15603

3          0.026            0.078       -0.491            0.24108

4          0.014             0.056       -0.513           0.26317

5          0.009          0.045       -0.524           0.27458

The mean =           0.569 (Sum of expected value)                          

Variance = mean of squared differences = 0.27665 (1.38323/5)

Standard deviation = square root of variance = 0.526 = 0.5

The mean, variance and standard deviation of the considered probability distribution are evaluated to be:

Supposing that the considered random variable is discrete, we get:

[tex]Mean = E(X) = \sum_{\forall x_i} f(x_i)x_i\\\\Variance = Var(X) = (\sum_{\forall x_i} f(x_i)x^2_i) - (E(X))^2\\[/tex]

As standard deviation is positive root of variance, thus,

[tex]\sigma = \sqrt {Var(X)}[/tex]

where [tex]x_i; \: \: i = 1,2, ... ,n[/tex] is its n data values

and  [tex]f(x_i)[/tex] is the probability of  [tex]X = x_i[/tex]

For this case, let we suppose that:

X = number of dogs per household in the considered town.

Then we have the probability distribution for values of X as:

Number of dogs     Probability

              0                  0.650

              1                   0.216

              2                  0.087

              3                  0.026

              4                  0.014

              5                  0.009

Thus, we get:

Mean of X : E(X) = [tex]0 \times 0.650 + 1 \times 0.216 + 2 \times 0087 + 3 \times 0.026 + 4 \times 0.014 + 5 \times 0.009 = 0.569[/tex]

Variance of X:

Var(X) =

[tex](0^2 \times 0.650 + 1^2 \times 0.216 + 2^2 \times 0.087 + 3^2 \times 0.026 + 4^2 \times (0.014)\\ + 5^2 \times (0.009)) - (0.569)^2 \approx 0.923[/tex]

Standard deviation of X = [tex]\sigma = \sqrt {Var(X)} \approx \sqrt{0.923} \approx 0.96[/tex]

Thus, the mean, variance and standard deviation of the considered probability distribution are evaluated to be:

Learn more about expectation of a random variable here:

https://brainly.com/question/4515179

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