http://www.ck12.org Chapter 7. Sampling Distributions and Estimations
The probability that the capacity attendance will be at least 12,000 is 0. 6 ( 0. 1 + 0. 1 + 0. 1 + 0. 1 + 0. 1 + 0. 1 )
Expected Values and Standard Deviation
Returning to the original problem of the number of cable hook-ups for the 500 single-family homes, take another
look at figures one and two. From these displays, you can find the mean number of hook-ups per household. You
expect 9.2% of households to have no hook-up, 32.8% to have one hook-up, 38.0% to have two hook-ups, 14.2% to
have three hook-ups and 5.8% to have four hook-ups. To calculate the mean number of hook-ups per household, use
the previous figure and add another column.
TABLE7.15:
Hook-ups per Household,x Proportion of Households,p Contribution to Mean,x·p
0 0. 092 0
1 0. 328 0. 328
2 0. 380 0. 760
3 0. 142 0. 426
4 0. 058 0. 232
Sum−→ 1. 746The mean of a probability distribution for the random variableXis denoted byμxorE(X)which representsexpected
value. Since you now know the expected number of cable hook-ups for each household, you can also calculate how
much each household will differ from this mean. In other words, you can calculate the expected standard deviation.
To do this, simply determine the expected value of the square of the deviations from the mean. As you recall from
chapter 1, this value is called the variance of the probability distribution, and gives a representation of how far an
actual value will in general stray from this mean.
σ^2 x=( 0 − 1. 746 )^2 ( 0. 092 )+( 1 − 1. 746 )^2 ( 0. 328 )+( 2 − 1. 746 )^2 ( 0. 380 )
+( 3 − 1. 746 )^2 ( 0. 142 )+( 4 − 1. 746 )^2 ( 0. 058 )
≈ 1. 0054The standard deviation (σx) is
√
- 0054 ≈ 1 .002. This indicates that each household will have 1.746 cable hook-ups
and differ from this mean by an average of about 1 hook-up. These calculations yield the following formulas for
calculating the expected value and the standard deviation (and its square, the variance) for a probability distribution.
E(X) =μx=∑xipi and Var(X) =σ^2 x=∑(xi−μx)^2 pi
σx=√
var(X)wherepiis the probability of the random variableXproduced whenxtakes on a specific valuexi.
Example:
Suppose an individual plays a gambling game where it is possible to lose $2.00, break even, win $6.00, or win
$20.00 each time he plays. The probability distribution for each outcome is provided by the following table: