Question

complete parts (a) and (b) below
the number of dogs per household in a small town
\begin{tabular}{lcccccc}dogs&0&1&2&3&4&5\probability&0.655&0.214&0.084&0.023&0.015&0.009end{tabular}
(a) find the mean, variance, and standard deviation of the probability distribution
find the mean of the probability distribution.
(mu = 0.6) (round to one decimal place as needed.)
find the variance of the probability distribution
(sigma^{2}=square) (round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for variance

The formula for variance \(\sigma^{2}=\sum(x - \mu)^{2}P(x)\), where \(x\) is the value of the random variable, \(\mu\) is the mean, and \(P(x)\) is the probability of \(x\).
We know \(\mu = 0.6\) (given).

Step2: Calculate \((x-\mu)^{2}P(x)\) for each \(x\)

• When \(x = 0\): \((0 - 0.6)^{2}\times0.655=( - 0.6)^{2}\times0.655 = 0.36\times0.655=0.2358\)
• When \(x = 1\): \((1 - 0.6)^{2}\times0.214=(0.4)^{2}\times0.214 = 0.16\times0.214 = 0.03424\)
• When \(x = 2\): \((2 - 0.6)^{2}\times0.084=(1.4)^{2}\times0.084=1.96\times0.084 = 0.16464\)
• When \(x = 3\): \((3 - 0.6)^{2}\times0.023=(2.4)^{2}\times0.023=5.76\times0.023 = 0.13248\)
• When \(x = 4\): \((4 - 0.6)^{2}\times0.015=(3.4)^{2}\times0.015 = 11.56\times0.015=0.1734\)
• When \(x = 5\): \((5 - 0.6)^{2}\times0.009=(4.4)^{2}\times0.009=19.36\times0.009 = 0.17424\)

Step3: Sum up the values

\(\sigma^{2}=0.2358+0.03424 + 0.16464+0.13248+0.1734+0.17424\)
\(\sigma^{2}=0.9148\approx0.9\)

Answer:

\(\sigma^{2}=0.9\)