Question

in the following probability distribution, the random variable x represents the number of activities a parent of a 6th - 8th grade student is involved in.

x	0	1	2	3	4
p(x)	0.059	0.108	0.248	0.303	0.282

a) compute and the mean of the random variable x.
b) compute the variance of the random variable x.
c) compute the standard deviation of the random variable x.
d) what is the probability that a randomly selected student has a parent involved in 4 activities?

Explanation:

Step1: Recall mean formula

The mean $\mu$ of a discrete - random variable $X$ is given by $\mu=\sum_{i}x_{i}P(x_{i})$. So, $\mu=(0\times0.059)+(1\times0.108)+(2\times0.248)+(3\times0.303)+(4\times0.282)$.

Step2: Calculate each product

$0\times0.059 = 0$, $1\times0.108=0.108$, $2\times0.248 = 0.496$, $3\times0.303=0.909$, $4\times0.282 = 1.128$.

Step3: Sum the products

$\mu=0 + 0.108+0.496 + 0.909+1.128=2.641$.

Step4: Recall variance formula

The variance $\sigma^{2}=\sum_{i}(x_{i}-\mu)^{2}P(x_{i})$.
$(0 - 2.641)^{2}\times0.059+(1 - 2.641)^{2}\times0.108+(2 - 2.641)^{2}\times0.248+(3 - 2.641)^{2}\times0.303+(4 - 2.641)^{2}\times0.282$.
$( - 2.641)^{2}\times0.059+( - 1.641)^{2}\times0.108+( - 0.641)^{2}\times0.248+(0.359)^{2}\times0.303+(1.359)^{2}\times0.282$.
$6.975881\times0.059 + 2.693881\times0.108+0.410881\times0.248+0.128881\times0.303+1.847881\times0.282$.
$0.411577+0.290939+0.101899+0.039051+0.521002 = 1.364468$.

Step5: Recall standard - deviation formula

The standard deviation $\sigma=\sqrt{\sigma^{2}}$. So, $\sigma=\sqrt{1.364468}\approx1.168$.

Answer:

b) $1.364468$
c) $1.168$