Question

chapter 3 - 4 practice test
due sun sep 23, 2025 11:59pm
attempt 1
in progress
next up: submit assignment
details
no additional details were added for this assignment.
chapter 3 - 4 practice test
25 points possible answered 6/25
question 1
consider the discrete random variable x given in the table below. calculate the mean, variance, and standard deviation of x.

x	11	12	13	17
p(x)	0.09	0.12	0.51	0.14	0.14

what is the expected value of x?

Explanation:

Step1: Calculate the mean ($\mu$)

The formula for the mean of a discrete - random variable is $\mu=\sum_{i}x_ip_i$.
$\mu=(10\times0.09)+(11\times0.12)+(12\times0.51)+(13\times0.14)+(17\times0.14)$
$=0.9 + 1.32+6.12 + 1.82+2.38$
$=12.54$

Step2: Calculate the variance ($\sigma^{2}$)

The formula for the variance is $\sigma^{2}=\sum_{i}(x_i - \mu)^2p_i$.
$(10 - 12.54)^2\times0.09+(11 - 12.54)^2\times0.12+(12 - 12.54)^2\times0.51+(13 - 12.54)^2\times0.14+(17 - 12.54)^2\times0.14$
$=(- 2.54)^2\times0.09+(-1.54)^2\times0.12+(-0.54)^2\times0.51+(0.46)^2\times0.14+(4.46)^2\times0.14$
$=6.4516\times0.09 + 2.3716\times0.12+0.2916\times0.51 + 0.2116\times0.14+19.8916\times0.14$
$=0.580644+0.284592 + 0.148716+0.029624+2.784824$
$=3.8284$

Step3: Calculate the standard deviation ($\sigma$)

The standard deviation is the square - root of the variance. $\sigma=\sqrt{\sigma^{2}}$.
$\sigma=\sqrt{3.8284}\approx1.9566$

Answer:

Mean: $12.54$
Variance: $3.8284$
Standard Deviation: $1.9566$
Expected Value (same as mean for discrete random variable): $12.54$