CK-12 Probability and Statistics - Advanced

http://www.ck12.org Chapter 4. Discrete Probability Distribution

μ=E(x) =∑

x

x p(x)

μ= 0 (. 002 )+ 1 (. 029 )+ 2 (. 132 )+ 3 (. 309 )+ 4 (. 360 )+ 5 (. 168 )

μ= 3. 50

b. We first calculate the variance ofx:

σ^2 =∑

x

(x−μ)^2 p(x)

=( 0 − 3. 5 )^2 (. 002 )+( 1 − 3. 5 )^2 (. 029 )+( 2 − 3. 5 )^2 (. 132 )

+( 3 − 3. 5 )^2 (. 309 )+( 4 − 3. 5 )^2 (. 36 )+( 5 − 3. 5 )^2 (. 168 )

σ^2 = 1. 05

Now we calculate the standard deviation,

σ=

√

σ^2 =

√

1. 05 = 1. 02

c. The graph ofp(x)is shown below.

We can use the meanμand the standard deviationσto describep(x)in the same way we used ̄xandsto describe the
relative frequency distribution. Notice thatμ= 3 .5 locates the center of the probability distribution. In other words,
if the five cancer patients receive chemotherapy treatment we expect the numberxthat are cured to be near 3.5. The
standard deviationσ= 1 .02 measures the spread of the probability distributionp(x).

Lesson Summary

1. Themean valueorexpected valueof a discrete random variablexis given byμ=E(x) =∑xx p(x).


2. Thevarianceof a discrete random variable is given byσ^2 =∑x(x−μ)^2 p(x).


3. The square root of the varianceσ^2 is thestandard deviationof a discrete random variable,σ=

√

σ^2.