Question

consider the probability distribution for the random variable x shown here. complete parts a through c below.

x	10	20	30	40	50	60

a. calculate \\( \mu \\), \\( \sigma^2 \\), and \\( \sigma \\)
\\( \mu = 33.5 \\) (type an integer or a decimal. do not round.)
\\( \sigma^2 = \square \\) (type an integer or a decimal. do not round.)

Explanation:

Step1: Recall Variance Formula

The formula for the variance \(\sigma^2\) of a discrete random variable is \(\sigma^2=\sum[(x - \mu)^2p(x)]\), where \(x\) are the values of the random variable, \(\mu\) is the mean, and \(p(x)\) are the corresponding probabilities. We know \(\mu = 33.5\), and the values of \(x\) are \(10,20,30,40,50,60\) with probabilities \(0.05,0.25,0.30,0.20,0.10,0.10\) respectively.

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

• For \(x = 10\):

\((10 - 33.5)^2\times0.05=(- 23.5)^2\times0.05 = 552.25\times0.05=27.6125\)

• For \(x = 20\):

\((20 - 33.5)^2\times0.25=(-13.5)^2\times0.25 = 182.25\times0.25 = 45.5625\)

• For \(x = 30\):

\((30 - 33.5)^2\times0.30=(-3.5)^2\times0.30=12.25\times0.30 = 3.675\)

• For \(x = 40\):

\((40 - 33.5)^2\times0.20=(6.5)^2\times0.20 = 42.25\times0.20=8.45\)

• For \(x = 50\):

\((50 - 33.5)^2\times0.10=(16.5)^2\times0.10 = 272.25\times0.10=27.225\)

• For \(x = 60\):

\((60 - 33.5)^2\times0.10=(26.5)^2\times0.10 = 702.25\times0.10 = 70.225\)

Step3: Sum the values

Now, sum all these results:
\(27.6125+45.5625 + 3.675+8.45+27.225+70.225\)
First, \(27.6125+45.5625=73.175\)
Then, \(73.175+3.675 = 76.85\)
Then, \(76.85+8.45=85.3\)
Then, \(85.3+27.225 = 112.525\)
Then, \(112.525+70.225=182.75\)

Answer:

\(182.75\)