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Probability and Statistics
The Binomial Probability Distribution
and Related Topics
Chapter 5
Section 1
Introduction to Random Variables and
Probability Distributions
Essential Question: How are the mean and the standard deviation determined from a
discrete probability distribution?
Student Objectives: The student will distinguish between random and discrete random
variables.
The student will graph discrete probability distributions.
The student will compute the mean and standard deviation for a
discrete probability distribution.
The student will compute the mean and standard deviation for a
linear function of a random variable x.
The student will compute the mean and standard deviation for a
linear combination of two independent random variables.
Terms:
Continuous random variable
Discrete random variable
Linear function of a random variable
Linear function of two independent random variables
Mean ( µ )
Probability Distribution
Random variable
Standard deviation (! )
Equations:
Mean: Expected value for a discrete probability distribution: µ = " ( x ! P ( x ))
Standard Deviation: ! =
$( x " µ )
2
# P( x)
Linear function: L = a + bx
Mean of a linear function: µ L = a + bµ
Variance of a linear function: ! L2 = b 2! 2
Standard deviation for a linear function: ! L = b 2! 2 = b !
Linear combination: W = ax1 + bx2
Mean of a linear combination: µW = aµ1 + bµ2
Variance of a linear combination: ! W2 = a 2! 12 + b 2! 22
Standard deviation for a linear function: ! W = a 2! 12 + b 2! 22
Graphing Calculator Skills:
Calculating the values of the mean and standard deviation for a discrete probability
distribution: 1-Var Stats L1 ,L 2 ! where L1 contains the x-values and L2 contains the
probabilities.
Sample Questions:
Directions: In problems #1 - 7, identify each of the following as either a discrete or continuous
random variable.
1. The number of people who are in a car.
2. The number of miles you drive in one week.
3. The weight of a box of cereal.
4. The number of boxes of cereal you buy in one year.
5. The length of time you spend eating your lunch.
6. The number of patients on a psychiatric ward in one day.
7. The volume of blood that is transfused during an operation.
8. In a personality inventory test for passive-aggressive traits, the possible scores are:
1 = extremely passive
2 = moderately passive
3 = neither
4 = moderately aggressive
5 = extremely aggressive
The test was administered to a group of 110 people and the results were as follows:
x (score)
1
2
3
4
5
f (frequency)
19
23
32
26
10
Construct a probability distribution table, calculate the expected value (the mean) and the
standard deviation. Use a histogram to graph the probability distribution.
x
f
1
19
2
23
3
32
4
26
5
10
Sum:
P(x)
x ! P( x)
x !µ
( x ! µ)
2
( x ! µ)
2
" P( x)
9. A local automotive repair shop has two work centers. The first center examines the car
with a computer diagnostic machine and the second center repairs the car. Let x1 and x2
be random variables representing the time in minutes to diagnose and repair the car.
Assume x1 and x2 are independent random variables. A recent study of the repair center
produced the following data:
Computer diagnose
Repair
x1 :
µ1 = 18.5 minutes
and
! 1 = 5.6 minutes
x2 : µ2 = 137.75 minutes and ! 2 = 17.2 minutes
a. Suppose it costs $1.25 per minute to diagnose the automobile and $0.95 per minute
to repair the automobile (without parts). Compute the linear combination
equation, mean, variance, and the standard deviation of W.
b. The repair shop charges a flat rate of $0.75 per minute to diagnose the car and if no
repairs are needed, there is an additional $25.00 service charge. Compute the
linear equation, mean, variance, and standard deviation of L.
Homework:
Pages 190 - 195
Exercises: #1 - 19 odd
Exercises: #2 - 18 even
ANSWERS
Directions: In problems #1 - 7, identify each of the following as either a discrete or continuous
random variable.
1. The number of people who are in a car.
DISCRETE
2. The number of miles you drive in one week.
CONTINUOUS
3. The weight of a box of cereal.
CONTINUOUS
4. The number of boxes of cereal you buy in one year.
DISCRETE
5. The length of time you spend eating your lunch.
CONTINUOUS
6. The number of patients on a psychiatric ward in one day.
DISCRETE
7. The volume of blood that is transfused during an operation.
CONTINUOUS
8. In a personality inventory test for passive-aggressive traits, the possible scores are:
1 = extremely passive
2 = moderately passive
3 = neither
4 = moderately aggressive
5 = extremely aggressive
The test was administered to a group of 110 people and the results were as follows:
x (score)
1
2
3
4
5
f (frequency)
19
23
32
26
10
Construct a probability distribution table, calculate the expected value (the mean) and the
standard deviation. Use a histogram to graph the probability distribution.
x
f
P(x)
x ! P( x)
x !µ
( x ! µ)
1
19
0.1727
0.1727
-1.8636
3.4731
0.5999
2
23
0.2091
0.4182
-0.8636
0.7459
0.1560
3
32
0.2909
0.8727
0.1364
0.0186
0.0054
4
26
0.2364
0.9455
1.1364
1.2913
0.3052
5
10
0.0909
0.4545
2.1364
4.5640
0.4149
Sum:
110
1.0000
2.8636
µ = 2.8636
2
( x ! µ)
2
" P( x)
1.4814
! = 1.4814
! = 1.2171
9. A local automotive repair shop has two work centers. The first center examines the car
with a computer diagnostic machine and the second center repairs the car. Let x1 and x2
be random variables representing the time in minutes to diagnose and repair the car.
Assume x1 and x2 are independent random variables. A recent study of the repair center
produced the following data:
Computer diagnose
Repair
x1 :
µ1 = 18.5 minutes
and
! 1 = 5.6 minutes
x2 : µ2 = 137.75 minutes and ! 2 = 17.2 minutes
a. Suppose it costs $1.25 per minute to diagnose the automobile and $0.95 per minute
to repair the automobile (without parts). Compute the linear combination
equation, mean, variance, and the standard deviation of W.
W = 1.25x1 + 0.95x2
µW = 1.25 (18.5 ) + 0.95 (137.75 )
µW = 23.125 + 130.8625
µW = 153.9875
µW = $153.99
! W2 = (1.25 ) ! 12 + ( 0.95 ) ! 22
2
! W = 315.9956
! W = 17.7763
! W = $17.78
2
! W2 = 1.5625! 12 + 0.9025! 22
! W2 = 1.5625 ( 5.6 ) + 0.9025 (17.2 )
2
2
! W2 = 1.5625 ( 31.36 ) + 0.9025 ( 295.84 )
! W2 = 49 + 266.9956
! W2 = 315.9956
! W2 = $316.00
b. The repair shop charges a flat rate of $0.75 per minute to diagnose the car and if no
repairs are needed, there is an additional $25.00 service charge. Compute the
linear equation, mean, variance, and standard deviation of L.
L = 0.75x1 + 25
! 2L = ( 0.75 ) ! 12
2
! 2L = 0.5625! 12
µ L = 25 + 0.75 (18.5 )
! 2L = 0.5625 ( 5.6 )
µ L = 25 + 13.875
µ L = 38.875
µ L = $38.88
! 2L = 17.64
2
! 2L = 0.5625 ( 31.36 )
! 2L = $17.64
! L = 17.64
! L = 4.2
! L = $4.20