7.1 – Discrete and Continuous Random Variables

7.1 – Discrete and Continuous Random Variables

Random Variable:

A variable whose value is a numerical outcome of a random phenomenon

Discrete Random Variable:

X has a countable number of possible values.

X

P(X)

x1 x2 x3 … xk

p1 p2 p3 …pk

Properties:

1. Every probability pi is between 0 and 1

2. The sum of the probabilities add to 1

Example #1:Make a probability histogram of the following grades on a four-point scale.

Grade 0 1 2 3 4

P(X) 0.05 0.28 0.19 0.32 0.16

Prob. Histogram of Grades

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5

Grade

Pro

bab

ilit

y

a. Determine if this is a legitimate probability distribution.

Grade 0 1 2 3 4

P(X) 0.05 0.28 0.19 0.32 0.16


Yes, 0.05 + 0.28 + 0.19 + 0.32 + 0.16 = 1And all individual probabilities are less than 1

b. P(X = 2)

Grade 0 1 2 3 4

P(X) 0.05 0.28 0.19 0.32 0.16


= 0.19

c. P(X 2) = 0.19 + 0.28 + 0.05 = 0.52

d. P(X < 2) = 0.28 + 0.05 = 0.33

e. P(X > 2) = 0.32 + 0.16 = 0.48

Continuous Random Variable:

X takes all values in an interval of numbers.

Probability distribution is a density curve, where the probability is the area under the curve.

Properties:

2. The height of the curve doesn’t represent the probability, the area does!

3. If it is normally distributed, then use the z-score,

1. Probability of an individual event, P(X = # ) = 0.

x -

Z =

Example #2Find the following probabilities for the Uniform distribution.

a. P(X > 0.2)

.1 .2 .3 .4 .5 .6 .7 .8 . 9 1

1.9.8.7.6.5.4.3.2.1

bh = (.8)(1) = 0.8


b. P(X = 0.3)

.1 .2 .3 .4 .5 .6 .7 .8 . 9 1

1.9.8.7.6.5.4.3.2.1

bh = (0)(1) = 0


c. P(0.1 < X < 0.7)

.1 .2 .3 .4 .5 .6 .7 .8 . 9 1

1.9.8.7.6.5.4.3.2.1

bh = (0.6)(1) = 0.6


d. P(0.6 < X < 1.0)

.1 .2 .3 .4 .5 .6 .7 .8 . 9 1

1.9.8.7.6.5.4.3.2.1

bh = (0.4)(1) = 0.4


e. P(0.6 X 1.0)

.1 .2 .3 .4 .5 .6 .7 .8 . 9 1

1.9.8.7.6.5.4.3.2.1

bh = (0.4)(1) = 0.4


f. What is different about continuous and discrete values according to letters d and e?

Continuous RV’s have a probability of zero when equal to one number, so the probability doesn’t change if less than or less than or equal to.

Example #3The amount of money spent by students for textbooks in a semester is a normally distributed random variable with a mean of $235 and a standard deviation of $15.

a. What percent of students spend less than $225 in a semester?

=235

= 15

x 225

Z = x – 225 – 235 15

=

-10 15=

-0.67=

P(Z < -0.67) =

OR: Normalcdf(lowerbound, upperbound, , )

Normalcdf(-100000, 225, 235, 15)

P(X < 225) = 0.2525


b. What percent of students spend more than $270 on textbooks in any semester?

Z = x – 270 – 235 15

=

3515=

2.33=

=235

= 15

x 270

P(Z > 2.33) =


Normalcdf(270, 1000000000, 235, 15)

P(X > 270) = 0.0098

1 – P(Z < 2.33) = 1 – 0.9901 = 0.0099


c. What is the probability that a student spends between $220 and $250 in any semester?

Z = x –

220 – 235 15

=

-15 15=

-1=

= 34 Z 250

Z 220

= 15Z = x –

250 – 235 15

=

1515=

1=

P(220 < X < 250) =


Normalcdf(220, 250, 235, 15)

P(220 < X < 250) = 0.6827

P(Z < 1) – P(Z < -1) = 0.8413 – 0.1587 = 0.6826

P(-1 < Z < 1) =

7.2 – Means and Variances of Random Variables

Mean of a Discrete Random Variable: (Expected Value)

To find the mean of the Random Variable X, multiply each possible outcome by its associated probability and sum all of the products.

Notation: X = X1p1 + X2p2 + X3p3 + … + Xkpk = Xipi

Example #1How much would you expect to win or lose per ticket in this California Decco Lottery Game on average (over the long run)?

# of matches

4 3 2 1 0

X = net gain

$4,999 $49 $4 0 -$1

Probability .000035 .00168 .0303 .242 .726

X = Xipi = (4,999x0.000035) + (49x0.00168) + (4x0.303) + (0x0.242) + (-1x0.726) =

-0.3475X =

Rules for Means:

a.If X is a random variable and a and b are fixed numbers, then a±bX = a ± bX

b. If X and Y are random variables, then x±y = X ± Y

Example #2Suppose the equation Y = 20 + 10X converts

a PSAT math score, X, into an SAT math score, Y. Suppose the average PSAT math score is 48. What is the average SAT math score?

20+100X = 20 + 10(48) = 20 + 480 = 500

Let X = 625 represent the average SAT math score. Let Y = 590 represent the average SAT verbal score. What are the averages combined?

X+Y = X + Y = 625 + 590 = 1215

Variance and Standard Deviation of the Discrete Random Variable

To find the variance, subtract the mean (expected value) from each observation, square it and multiply it by its associated probability. The sum of these results is the variance. The Standard Deviation is simply the square root of the variance. They must be independent to use the rule!!!

Variance:

2 2 2 221 1 2 2 3 3 ...x x x x k x kx p x p x p x p

22x i x ix p

Standard Deviation:

22x x i x ix p

Example #3Find the variance and the standard deviation.

# of matches

4 3 2 1 0

X = net gain

$4,999 $49 $4 0 -$1

Probability .000035 .00168 .0303 .242 .726

2 21( )x z ix p

2(4999 ( .3475)) (0.000035) 2(4 ( .3475)) (0.0303) 2(49 ( .3475)) (0.00168)

2 2(0 ( .3475)) (0.242) ( 1 ( .3475)) (0.726)

879.7606 29.6608

Rules for Variances:If X is a random variable and a and b are fixed numbers, then XbXa b 2 2 2

a bX Xb

If X and Y are independent random variables, then:

X Y2 x

2 Y2

X Y2 x

2 Y2

and

2YXYX Therefore: and

2X Y X Y

Example #4Consider a distribution with μX = 14, σX = 2. Multiply each value of X by 4 and then add 3 to each. Find the mean and standard deviation.

3 4X

3 4X

3 4(14) 59

4(2) 8

Example #4You have two scales for measuring weights in a chemistry lab. Both scales give answers that vary a bit in repeated weighings of the same item. If the true weight of a compound is 2.00 grams (g), the first scale produces reading X that have mean 2.00g and standard deviation 0.002g. The second scale’s readings Y have mean 2.001g and standard deviation 0.001g. What are the mean and standard deviation of the difference Y – X between the readings? (The readings are independent)

Y – X =

Y =X = 2.00g

X = Y =0.002g

2.001g

0.001g

2.001 – 2.00 = 0.001

Y =X = 2.00g

X = Y =0.002g

2.001g

0.001g

Y – X = 2 2Y X 2 2

0.001 0.002

.000005 0.0022

Example #5Rotter Partners is planning a major investment. The amount of profit X is uncertain, but a probabilistic estimate gives the following distribution (in millions of dollars):

Profit: 1 1.5 2.0 4 10

Probability: 0.1 0.2 0.4 0.2

a. Find the missing probability.

0.1


Profit: 1 1.5 2.0 4 10

Probability: 0.1 0.2 0.4 0.2 0.1

b. Find the mean and standard deviation of the profit

X = Xipi = (1x0.1) + (1.5x0.2) + (2x0.4) + (4x0.2) + (10x0.1)=

3X =

2(1 3) (0.1)

21( )X z ix p

2(2 3) (0.4)2(1.5 3) (0.2) 2 2(4 3) (0.2) (10 3) (0.1)

6.35 2.5199


Profit: 1 1.5 2.0 4 10

Probability: 0.1 0.2 0.4 0.2 0.1

b. Find the mean and standard deviation of the profit

c. Rotter Partners owes its source of capital a fee of $200,000 plus 10% of the profits X. So, the firm actually retains:

Y = 0.9X – 0.2

from the investment. Find the mean and standard deviation of Y.

3X = 2.5199X

0.9X – 0.2 = 0.9(3) – 0.2

Y = 2.5

Y = Y =

0.9(2.5199)

0.9X – 0.2 =

Y =

2.2691Y =

Calculator Tip: Expected Value And SD

Stat – Calc – 1-VarStats – L1, L2

Example #6Typographical errors can be either “nonword errors” or “word errors.” A nonword error is not a real word, as when “the” is typed “teh.” A word error is a real word, but not the right word, as when “lose” is typed as “loose.” When undergraduate students are asked to write a 250-word essay (without spell-checking):

The number of nonword errors has the following distribution:

Errors 0 1 2 3 4

Probability 0.1 0.2 0.3 0.3 0.1

The number of word errors has this distribution:

Errors 0 1 2 3

Probability 0.4 0.3 0.2 0.1

Find the mean and standard deviation of the total number of errors in an essay if the counts of word errors and nonword errors are independent.

Word Errors Non-Word Errors

W = 2.1

W = 1.1358

N = 1

N = 1

2.1 + 1 = 3.1W+N = W+ N =

W+N = 2 2W N 2 2

1.1358 1 2.29 1.5133

Documents

7.1 – Discrete and Continuous Random Variables