complete parts (a) and (b) below.
the number of dogs per household in a small town
\begin{array}{lcccccc}
\text{dogs} & 0 & 1 & 2 & 3 & 4 & 5 \\
\text{probability} & 0.676 & 0.203 & 0.078 & 0.023 & 0.012 & 0.007
\end{array}

(a) find the mean, variance, and standard deviation of the probability distribution.
find the mean of the probability distribution.
$\mu = 0.5$ (round to one decimal place as needed.)
find the variance of the probability distribution.
$\sigma^2 = $ (round to one decimal place as needed.)

Question

complete parts (a) and (b) below.
the number of dogs per household in a small town
\begin{array}{lcccccc}
\text{dogs} & 0 & 1 & 2 & 3 & 4 & 5 \\
\text{probability} & 0.676 & 0.203 & 0.078 & 0.023 & 0.012 & 0.007
\end{array}

(a) find the mean, variance, and standard deviation of the probability distribution.
find the mean of the probability distribution.
$\mu = 0.5$ (round to one decimal place as needed.)
find the variance of the probability distribution.
$\sigma^2 = $ (round to one decimal place as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Mean of Probability Distribution" ], "new_concepts": [ "Variance of Probability Distribution" ], "current_concepts": [ "Mean of Probability Distribution", "Variance of Probability Distribution" ] } ## Calculate the mean of the probability distribution $$ \begin{aligned} \mu &= \sum x \cdot P(x) \ &= 0(0.676) + 1(0.203) + 2(0.078) + 3(0.023) + 4(0.012) + 5(0.007) \ &= 0 + 0.203 + 0.156 + 0.069 + 0.048 + 0.035 \ &= 0.511 \approx 0.5 \end{aligned} $$ ## Calculate the variance of the probability distribution $$ \begin{aligned} \sigma^2 &= \sum x^2 \cdot P(x) - \mu^2 \ \sum x^2 \cdot P(x) &= 0^2(0.676) + 1^2(0.203) + 2^2(0.078) + 3^2(0.023) + 4^2(0.012) + 5^2(0.007) \ &= 0 + 0.203 + 4(0.078) + 9(0.023) + 16(0.012) + 25(0.007) \ &= 0.203 + 0.312 + 0.207 + 0.192 + 0.175 \ &= 1.089 \ \sigma^2 &= 1.089 - (0.511)^2 \ &= 1.089 - 0.261121 \ &= 0.827879 \approx 0.8 \end{aligned} $$ Find the variance of the probability distribution. $\sigma^2 =$ $0.8$ (Round to one decimal place as needed.) { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Variance of Probability Distribution" ] }

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Calculate the mean of the probability distribution

Calculate the variance of the probability distribution

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