use the probability distribution to complete parts (a) and (b) below.

the number of school-related extracurricular activities per student
$\begin{array}{lcccccccc}\text{activities} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text{probability} & 0.059 & 0.122 & 0.162 & 0.179 & 0.214 & 0.127 & 0.086 & 0.051\end{array}$

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

the mean is 3.3.
(round to one decimal place as needed.)

the variance is .
(round to one decimal place as needed.)

Question

use the probability distribution to complete parts (a) and (b) below.

the number of school-related extracurricular activities per student
$\begin{array}{lcccccccc}\text{activities} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text{probability} & 0.059 & 0.122 & 0.162 & 0.179 & 0.214 & 0.127 & 0.086 & 0.051\end{array}$

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

the mean is 3.3.
(round to one decimal place as needed.)

the variance is  .
(round to one decimal place as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Mean of Probability Distribution", "Variance of Probability Distribution" ], "new_concepts": [], "current_concepts": [ "Mean of Probability Distribution", "Variance of Probability Distribution", "Discrete Probability Distribution" ] } ## Calculate the mean of the probability distribution $$ \begin{aligned} \mu = \sum x \cdot P(x) &= 0(0.059) + 1(0.122) + 2(0.162) + 3(0.179) + 4(0.214) + 5(0.127) + 6(0.086) + 7(0.051) \ &= 0 + 0.122 + 0.324 + 0.537 + 0.856 + 0.635 + 0.516 + 0.357 \ &= 3.347 \approx 3.3 \end{aligned} $$ ## Calculate the variance of the probability distribution $$ \begin{aligned} \sigma^2 = \sum x^2 \cdot P(x) - \mu^2 \ \sum x^2 \cdot P(x) &= 0^2(0.059) + 1^2(0.122) + 2^2(0.162) + 3^2(0.179) + 4^2(0.214) + 5^2(0.127) + 6^2(0.086) + 7^2(0.051) \ &= 0 + 0.122 + 0.648 + 1.611 + 3.424 + 3.175 + 3.096 + 2.499 \ &= 14.575 \ \sigma^2 &= 14.575 - (3.347)^2 \ &= 14.575 - 11.202409 \ &= 3.372591 \approx 3.4 \end{aligned} $$ The variance is 3.4 { "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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