Week 1 - Sums and Differences of Independent Random Variables Flashcards
This is the set of values that a random variable can take on
Probability distribution
there are three ways
that you can create probability distributions from data. Sometimes previously collected data, relative to the random
variable that you are studying, can help to create a probability distribution. In addition to this method, a simulation is
also a good way to create an approximate probability distribution
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A probability distribution can also be constructed
from the basic principles, assumptions, and rules of theoretical probability. The examples in this lesson will lead
you to a better understanding of these rules of theoretical probability.
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Example: Create a table that shows all the possible outcomes when two dice are rolled simultaneously. (Hint: There
are 36 possible outcomes.)
*Table 4.5 on Module
This table of possible outcomes when two dice are rolled simultaneously that is shown above can now be used to
construct various probability distributions. The first table below displays the probabilities for all the possible sums
of the two dice, and the second table shows the probabilities for each of the possible results for the larger of the two
numbers produced by the dice.
Table 4.6 on Module
When you roll the two dice, what is the probability that the sum is 4? By looking at the first table above, you can
see that the probability is 3
36 .
What is the probability that the larger number is 4? By looking at the second table above, you can see that the
probability is 7
36 .
Example: The Regional Hospital has recently opened a new pulmonary unit and has released the following data on
the proportion of silicosis cases caused by working in the coal mines. Suppose two silicosis patients are randomly
selected from a large population with the disease.
Silicosis Cases Proportion
Worked in the mine 0.80
Did not work in the mine 0.20
There are four possible outcomes for the two patients. With ’yes’ representing “worked in the mines” and ’no’
representing “did not work in the mines”, the possibilities are as follows:
Table 4.9 on module
As stated previously, the patients for this survey have been randomly selected from a large population, and therefore the outcomes are independent. The probability for each outcome can be calculated by multiplying the appropriate
proportions as shown:
Solution on Module
If X represents the number silicosis patients who worked in the mines in this random sample, then the first of these
outcomes results in x = 0, the second and third each result in x = 1, and the fourth results in x = 2. Because the
second and third outcomes are disjoint, their probabilities can be added. The probability distribution for X is given
in the table below:
Table 4.10 on Module
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:
Table 4.11 on Module
The table can be used to calculate the expected value and the variance of this distribution:
Solution on module
Thus, the player can expect to win $2.60 playing this game.
The variance of this distribution can be calculated as shown:
Solution on module
Example: The following probability distribution was constructed from the results of a survey at the local university.
The random variable is the number of fast food meals purchased by a student during the preceding year (12 months).
For this distribution, calculate the expected value and the standard deviation.
Table 4.12 on module
You must begin by estimating a mean for each interval, and this can be done by finding the center of each interval.
For the first interval of [1 6), 6 is not included in the interval, so a value of 3 would be the center. This same
procedure can be used to estimate the mean of all the intervals. Therefore, the expected value can be calculated as
follows:
Solution on module
Likewise, the standard deviation can be calculated:
Solution on module
Thus, the expected number of fast food meals purchased by a student at the local university is 13.93, and the standard
deviation is 14.43. Note that the mean should not be rounded, since it does not have to be one of the values in the
distribution. You should also notice that the standard deviation is very close to the expected value. This means that
the distribution will be skewed to the right and have a long tail toward the larger numbers.
Technology Note: Calculating mean and variance for probability distribution on TI-83/84 Calculator
Figure on module
Notice that the mean, which is denoted by x in this case, is 13.93, and the standard deviation, which is denoted by
sx, is approximately 14.43.
If you add the same value to all the numbers of a data set, the shape and standard deviation of the data set remain
the same, but the value is added to the mean. This is referred to as ______ the data set
Re-centering
Likewise, if you ______
the data, or multiply all the data values by the same nonzero number, the basic shape will not change, but the mean
and the standard deviation will each be a multiple of this number. (Note that the standard deviation must actually be
multiplied by the absolute value of the number.)
Rescale
If you multiply the numbers of a data set by a constant d and then
add a constant c, the mean and the standard deviation of the transformed values are expressed as follows:
Formula on module
These are called linear transformations, and the implications of this can be better understood if you return to the
casino example.
Example: The casino has decided to triple the prizes for the game being played. What are the expected winnings for
a person who plays one game? What is the standard deviation? Recall that the expected value was $2.60, and the
standard deviation was $6.45.
Solution:
The simplest way to calculate the expected value of the tripled prize is (3)($2.60), or $7.80, with a standard deviation
of (3)($6.45), or $19.35. Here, c = 0 and d = 3. Another method of calculating the expected value and standard
deviation would be to create a new table for the tripled prize:
Table 4.13 on module
The calculations can be done using the formulas or by using a graphing calculator. Notice that the results are the
same either way.
This same problem can be changed again in order to introduce the Addition Rule and the Subtraction Rule for random
variables. Suppose the casino wants to encourage customers to play more, so it begins demanding that customers
play the game in sets of three. What are the expected value (total winnings) and standard deviation now?
Let X,Y and Z represent the total winnings on each game played. If this is the case, then µX+Y+Z is the expected
value of the total winnings when three games are played. The expected value of the total winnings for playing one
game was $2.60, so for three games the expected value is:
Figure on Module
Thus, the expected value is the same as that for the tripled prize.
Since the winnings on the three games played are independent, the standard deviation of X,Y and Z can be calculated
as shown below:
Figure on module
This means that the person playing the three games can expect to win $7.80 with a standard deviation of $11.17.
Note that when the prize was tripled, there was a greater standard deviation ($19.36) than when the person played
three games ($11.17).
The Addition and Subtraction Rules for random variables are as follows:
If X and Y are random variables, then:
µX+Y = µX +µY
µXY = µX µY
If X and Y are independent, then:
Formula on module
Variances are added for both the sum and difference of two independent random variables, because the variation in
each variable contributes to the overall variation in both cases. (Subtracting is the same as adding the opposite.)
Suppose you have two dice, one die, X, with the usual positive numbers 1 through 6, and another, Y, with the
negative numbers 1 through 6. Next, suppose you perform two experiments. In the first, you roll the first die, X,
and then the second die, Y, and you compute the difference of the two rolls. In the second experiment, you roll the
first die and then the second die, and you calculate the sum of the two rolls.
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Solution on module
Example: Beth earns $25.00 an hour for tutoring but spends $20.00 an hour for piano lessons. She saves the
difference between her earnings for tutoring and the cost of the piano lessons. The numbers of hours she spends on
each activity in one week vary independently according to the probability distributions shown below. Determine her
expected weekly savings and the standard deviation of these savings.
Table 4.14 and Table 4.15 on module
X represents the number of hours per week taking piano lessons, and Y represents the number of hours tutoring per
week. The mean and standard deviation for each can be calculated as follows:
solution on module
The expected number of hours Beth spends on piano lessons is 1.1 with a standard deviation of 0.831 hours.
Likewise, the expected number of hours Beth spends tutoring is 2.6 with a standard deviation of 1.11 hours.
Beth spends $20 for each hour of piano lessons, so her mean weekly cost for piano lessons can be calculated with
the Linear Transformation Rule as shown:
µ20X = (20)(µX )=(20)(1.1) = $22 by the Linear Transformation Rule.
Beth earns $25 for each hour of tutoring, so her mean weekly earnings from tutoring are as follows:
µ25Y = (25)(µY )=(25)(2.6) = $65 by the Linear Transformation Rule.
Thus, Beth’s expected weekly savings are:
µ25Y µ20X = $65$22 = $43 by the Subtraction Rule.
The standard deviation of the cost of her piano lessons is:
s20X = (20)(0.831) = $16.62 by the Linear Transformation Rule.
The standard deviation of her earnings from tutoring is:
s25Y = (25)(1.11) = $27.75 by the Linear Transformation Rule.
Finally, the variance and standard deviation of her weekly savings is:
Solution on Module
