Question

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

the number of school-related extracurricular activities per student
activities: 0, 1, 2, 3, 4, 5, 6, 7
probability: 0.059, 0.122, 0.162, 0.179, 0.214, 0.127, 0.086, 0.051

(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 3.4.
(round to one decimal place as needed.)

the standard deviation is 1.8.
(round to one decimal place as needed.)

(b) interpret the results.

the mean is 3.3, so the average student is involved in activities. the standard deviation is 1.8, so the typical number of activities per student deviates from the mean by about 2 activities.
(round to one decimal place as needed.)

dropdown options:
3 or 4 activities.
at least 5 activities.
no activities.

Explanation:

Calculate the mean of the probability distribution

$$ \mu = \sum x \cdot P(x) = 0(0.059) + 1(0.122) + 2(0.162) + 3(0.179) + 4(0.214) + 5(0.127) + 6(0.086) + 7(0.051) = 3.272 \approx 3.3 $$

Interpret the mean in context

The mean is \(3.3\), which represents the average number of activities a student is involved in. Since \(3.3\) is between \(3\) and \(4\), the average student is involved in "3 or 4 activities."

Interpret the standard deviation in context

The standard deviation is \(1.8\), which measures the typical deviation from the mean. Rounded to the nearest whole number, \(1.8 \approx 2\), meaning the typical number of activities deviates from the mean by about \(2\) activities.

Answer:

The mean is 3.3, so the average student is involved in <blank>3 or 4 activities.</blank> The standard deviation is 1.8, so the typical number of activities per student deviates from the mean by about 2 activities.