Download Probability distribution Mean, Variance and Exception of random

Probability distribution
Mean, Variance and Exception of random variable
1-Mean
The formula for the mean of a probability distribution
The mean of random variable of a probability distribution is:
Where X1, X2, X3, . . . , Xn, are the outcomes and P(X1 ), P(X2 ), P(X3),….,P(Xn) are
the corresponding probabilities.
Note: ∑ X . P(X) means to sum the products.
Example -1Find the mean of the number of spots that appear when a die is tossed?
Solution
In the toss of a die, the mean can be computed as follows.
Example -2In a family with two children, find the mean of the number of children who
will be girls.
Solution
The probability distribution is as follows:
Example -3If three coins are tossed, find the mean number of heads that will occur.
Solution
The probability distribution is as follows:
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Example -4In a restaurant, the following probability distribution was obtained for the
number of items a person ordered for a large pizza.
Find the mean for the distribution.
Solution
2-Variance
Formula for the Variance of a Probability Distribution
Find the variance of a probability distribution by multiplying the square of
each outcome by its corresponding probability, summing those products, and
subtracting the square of the mean. The formula for the variance of a probability
distribution is:
The standard deviation of a probability distribution is
Not: Remember that the variance and standard deviation cannot be
negative.
Example -5Compute the variance and standard deviation for the data in Example (1).
Solution
Recall that the mean is µ = 3.5, as computed in Example (1). Square each
outcome and multiply by the corresponding probability, sum those products, and
then subtract the square of the mean.
To get the standard deviation, find the square root of the variance.
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Example -6Five balls numbered 0, 2, 4, 6, and 8 are placed in a bag. After the balls are
mixed, one is selected, its number is noted, and then it is replaced. If this
experiment is repeated many times, find the variance and standard deviation of
the numbers on the balls.
Solution
Let X be the number on each ball. The probability distribution is as
follows:
The mean, variance, and standard deviation can also be found by using
vertical columns, as shown. (0.2 is used for P(X) since 1/5 = 0.2)
X
P ( X)
0
2
4
6
8
0.2
0.2
0.2
0.2
0.2
X • P( X)
X2 • P(X)
0
0
0.4
0.8
1.2
1.6
∑ X• P(X) = 4
0.8
3.2
7.2
12.8
∑ X 2• P(X) = 24
Find the mean by summing the X • P(X) column and the variance by
summing the X2 • P(X) column and subtracting the square of the mean
Example -7The probability that 0, 1, 2, 3, or 4 people will be placed on hold when
they call a radio talk show is shown in the distribution. Find the variance and
standard deviation for the data. The radio station has four phone lines.
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Solution
3-Expectation
Related to the concept of the mean for a probability distribution is the
concept of expected value or expectation. Expected value is used in various types
of games of chance, in insurance, and in other areas, such as decision theory.
The expected value of a discrete random variable of a probability
distribution is the theoretical average of the variable. The formula is
The symbol E(X) is used for the expected value.
The formula for the expected value is the same as the formula for the
theoretical mean. The expected value, then, is the theoretical mean of the
probability distribution. That is, E(X) = µ.
When expected value problems involve money, it is customary to round the
answer to the nearest cent.
Example -8One thousand tickets are sold at $1 each for a color television valued at
$350. What is the expected value of the gain if a person purchases one ticket?
Solution
The problem can be set up as follows:
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Two things should be noted. First, for a win, the net gain is $349, since the
person does not get the cost of the ticket ($1) back. Second, for a loss, the gain is
represented by a negative number, in this case -$1. The solution, then, is
Expected value problems of this type can also be solved by finding the
overall gain and subtracting the cost of the tickets, as shown:
Here, the overall gain ($350) must be used.
Note that the expectation is —$0.65. This does not mean that a person
loses $0.65, since the person can only win a television set valued at $350 or lose
$1 on the ticket. What this expectation means is that the average of the losses is
$0.65 for each of the 1000 ticket holders. Here is another way of looking at this
situation: If a person purchased one ticket each week over a long period of time,
the average loss would be $0.65 per ticket, since theoretically, on average, that
person would win the set once for each 1000 tickets purchased.
Example -9A ski resort loses $70,000 per season when it does not snow very much and
makes $250,000 profit when it does snow a lot. The probability of it snowing at
least 75 inches (i.e., a good season) is 40%. Find the expectation for the profit.
Solution
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Example -10One thousand tickets are sold at $1 each for four prizes of $100, $50, $25,
and $10. What is the expected value if a person purchases two tickets?
Solution
H.W:
1- From past experience, a company has found that in cartons of transistors, 92%
contain no defective transistors, 3% contain one defective transistor, 3% contain
two defective transistors and 2% contain three defective transistors. Find the
mean, variance, and standard deviation for the defective transistors ?
2-A recent survey by an insurance company showed the following probabilities
for the number of automobiles each policyholder owned. Find the mean.
variance, and standard deviation for the distribution?
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