Question

consider the probability distribution shown for the random variable x found below. complete part a through f.

x | 5 | 6 | 8 | 14
p(x) | 0.4 | 0.2 | 0.2 | 0.2

a. find \\( \mu = e(x) \\).
\\( \mu = 7.6 \\) (round to the nearest tenth as needed )

b. find \\( \sigma^2 = e\left(x - \mu)^2\
ight \\).
\\( \sigma^2 = \square \\) (round to the nearest hundredth as needed )

Explanation:

Step1: Recall the formula for variance

The formula for the variance \(\sigma^2\) of a discrete random variable is \(\sigma^2 = E[(x - \mu)^2]=\sum (x - \mu)^2p(x)\), where \(\mu\) is the expected value, \(x\) are the values of the random variable, and \(p(x)\) are their corresponding probabilities. We know \(\mu = 7.6\), \(x\) values are 5, 6, 8, 14 and \(p(x)\) values are 0.4, 0.2, 0.2, 0.2 respectively.

Step2: Calculate \((x - \mu)^2\) for each \(x\)

• For \(x = 5\): \((5 - 7.6)^2=(- 2.6)^2 = 6.76\)
• For \(x = 6\): \((6 - 7.6)^2=(-1.6)^2 = 2.56\)
• For \(x = 8\): \((8 - 7.6)^2=(0.4)^2 = 0.16\)
• For \(x = 14\): \((14 - 7.6)^2=(6.4)^2 = 40.96\)

Step3: Multiply each \((x - \mu)^2\) by \(p(x)\) and sum

• For \(x = 5\): \(6.76\times0.4 = 2.704\)
• For \(x = 6\): \(2.56\times0.2 = 0.512\)
• For \(x = 8\): \(0.16\times0.2 = 0.032\)
• For \(x = 14\): \(40.96\times0.2 = 8.192\)

Now sum these values: \(2.704+0.512 + 0.032+8.192=11.44\)

Answer:

\(11.44\)