Download Standard Deviation

Lesson 7 - 2
Means and Variances
of Random Variables
Knowledge Objectives
• Define what is meant by the mean of a random
variable
• Explain what is meant by a probability distribution
• Explain what is meant by a uniform distribution
• Discuss the shape of a linear combination of
independent Normal random variables
Construction Objectives
• Calculate the mean of a discrete random variable.
• Calculate the variance and standard deviation of a
discrete random variable.
• Explain, and illustrate with an example, what is
meant by the law of large numbers.
• Explain what is meant by the law of small numbers.
• Given µx and µy, calculate µa+bx, and µx+y.
• Given x and y, calculate 2a+bx, and 2x+y (where x
and y are independent).
• Explain how standard deviations are calculated
when combining random variables.
Vocabulary
• Mean – balance point of the probability histogram or
density curve. Symbol: μx
• Standard Deviation – square root of the variance.
Symbol: x
• Variance – is the average squared deviation of the
values of the variable from their mean. Symbol: σ²x
Means
• Mean of a random variable is also known as its
expected value
Discrete Random Variable - Mean
The mean, or expected value [E(x)], of a discrete random
variable is given by the formula
μx = ∑ [x ∙P(x)]
where x is the value of the random variable and P(x)
is the probability of observing x
Mean of a Discrete Random Variable Interpretation:
If we run an experiment over and over again, the law
of large numbers helps us conclude that the difference
between x and ux gets closer to 0 as n (number of
repetitions) increases
Variance
Discrete Random Variable - Variance
Variance and Standard Deviation of a Discrete Random
Variable:
The variance of a discrete random variable is given by:
σ2x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x2 ∙ P(x)] – μ2x
and standard deviation is √σ2
Note: round the mean, variance and standard deviation to
one more decimal place than the values of the random
variable
Uniform PDF
An experiment is said to be a Uniform experiment
provided:
1. The probability of each value of the random variable
is equal (like in a six-sided die)
2. The trials are independent of each other (what
happened last does not affect what happens next)
Uniform PDF
If X is a value of the uniform random variable, then
probability formula for X is
1
P(x) = ------x = 0, 1, 2, 3, … , n
n
where n is the total number of discrete values of the
random variable x
Mean: μx = ∑ [x ∙P(x)] = (1/n)∑ x
Standard Deviation:
σ2x = ∑ [(x – μx)2 ∙ P(x)] = (1/n) ∑ [(x – μx)2
= ∑[x2 ∙ P(x)] – μ2x = (1/n) ∑ [x2 ] – μx2
Using your TI-83 calculator
We can use 1-Var-Stats to calculate the mean
and standard deviation of a discrete random
variable given it’s outcomes and probability
• Type in outcomes in L1
• Type in corresponding probabilities in L2
• Use 1-Var-Stats L1, L2 to get statistics
Example 1
You have a fair 10-sided die with the number 1 to 10 on
each of the faces.
Determine the mean and standard deviation.
Mean: ∑ [x ∙P(x)] = (1/10) (∑ x) = (1/10)(55) = 5.5
Var: ∑[x2 ∙ P(x)] – μ2x = (1/n) ∑ [x2 ] – μx2
= (1/10) (385) - 30.25)
= (38.5 – 30.25)
= 8.25
St Dev = 2.8723
Example 2
Below is a distribution for number of visits to a dentist
in one year.
X = # of visits to a dentist
x
0
1
2
3
4
P(x)
.1
.3
.4
.15
.05
Determine the expected value, variance and standard
deviation.
Mean: ∑ [x ∙P(x)] = (.1)(0) + (.3)(1) + (.4)(2) + (.15)(3) + (.05)(4)
= 0 + .3 + .8 + .45 +.2 = 1.75
Var: ∑[x2 ∙ P(x)] – μ2x = ∑ [x2 ∙ P(x)] – μx2
= (0 + .3 + .4(4) + .15(9) + .05(16) ) – 3.0625)
= 4.05 – 3.8626
= 0.9875
St Dev
= 0.9937
Example 3
What is the average size of an American family? Here
is the distribution of family size according to the
1990 Census:
# in family 2
p(x)
.413
3
.236
4
.211
5
.090
6
.032
Mean: ∑ [x ∙P(x)] = (.413)(2) + (.236)(3) + (.211)(4) +
(.09)(5) + (.032)(6) + (.018)(7)
= .826 + .708 + .844 + .45 + .192 + .126
= 3.146
7
.018
Example 4
You are trying to decide whether to take out a $250
deductible which will cost you $90 per year. Records
show that for this community the average cost of repair
is $900. Records also show that 10% of the drivers
have an accident during the year. If you have sufficient
assets so that you will not be financially handicapped if
you had to pay out the $900 or more for repairs, should
you buy the policy?
P(accident) = 0.1 so P(no accident) = 0.9
ave repair cost = $900
Yearly cost = $90
Expected Yearly with Policy: ∑ [x ∙P(x)] = (.1)(250) + (.9)(0) + 90
= 25 + 90
= $115
Expected Yearly without: ∑ [x ∙P(x)] = (.1)(900) + (.9)(0)
= $90
Summary and Homework
• Summary
– Expected value is the mean ∑ [x ∙P(x)]
– Variance is ∑[x2 ∙ P(x)] – μ2x
– Standard Deviation is variance
• Homework
– pg 486-7; 7.24 – 7.30