Business Statistics Online Test
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bstatsmodule4slides.pptxModule 4 Discrete Random variables
Master for Business Statistics
Dane McGuckian
Topics
4.1 Probability Distributions for Discrete Random Variables
4.2 Expected Value, Variance, and Standard Deviation for Discrete Random Variables
4.3 The Binomial Probability Distribution
4.4 The Poisson Probability Distribution
4.1
Probability Distributions for Discrete Random Variables
Discrete Random Variable
A discrete random variable is a variable that can only assume a countable number of values.
The achievable values of a discrete random variable are separated by gaps.
Example: a publisher may sell 300,000 or 300,001 copies of its latest book, but it cannot sell 300,000.159 copies of its latest book
Discrete random variables contain observations that are not measured on a continuous scale
Most often a discrete random variable contains observations that are derived from counting something.
Discrete Random Variables
Examples of Discrete Random variables:
The number of clicks received by an online advertisement over the past hour
The number of books sold by an author yesterday
The number of people missing the most recent flight from Miami to London
The number of parking violations last semester on campus
Discrete Probability Distributions
The probability distribution of a discrete random variables lists all of the possible outcomes for the random variable and the associated probability for each of those outcomes.
The distribution can be represented by a table, a graph or a formula.
| Number of Female Jurors, X | Probability of Outcome P(X) |
| 0 | 0.008 |
| 1 | 0.061 |
| 2 | 0.186 |
| 3 | 0.303 |
| 4 | 0.278 |
| 5 | 0.136 |
| 6 | 0.028 |
Characteristics of a Probability Distribution
A probability distribution lists all possible outcomes for the experiment and the corresponding probability for each of those outcomes.
Remember that the probabilities cannot be negative
Each probability must lie between zero and one
The sum of the probabilities for all of the outcomes must be one.
Example: Probability distribution of the number of free throws made by basketball players who make free throws 80% of the time (X). For instance, there is a 4%
chance that a player misses both throws
All these probabilities are non-negative
Each probability lies between 0 and 1
The sum of these probabilities are:
0.04+0.32+0.64 = 1
| X | P(X) |
| 0 | 0.04 |
| 1 | 0.32 |
| 2 | 0.64 |
4.2
Expected Value, Variance, and Standard Deviation for Discrete Random Variables
The Mean of a Discrete Probability Distribution
The average value for a probability distribution is referred to as the expected value of the probability distribution.
It represents the long-run typical value for the random variable.
If it were possible to run the trials indefinitely, the expected value would be the mean for the infinite set of outcomes for the random variable that would result from those trials.
The Expected value
The expected value is essentially a weighted average of the possible outcomes for the random variable. The weights are the corresponding probabilities for those outcomes.
Just as the arithmetic mean we studied earlier, it is common for the expected value to be a decimal of a fraction even when the original set of outcomes must be whole numbers.
The Expected value of a Discrete Random Variable
How much money on average will an insurance company make off of a 1-year life insurance policy worth $50,000, if they charge $1000.00 for the policy and each policy holder has a 0.9999 of surviving the year?
Average implies mean, and that mean “expected value” in the context of a probability distribution. The formula is:
If a person lives, the company makes
$1000 (hence it is “positive”); if the person
dies, it pays the family $50,000 (but also
gets $1000 from the family), so their loss
is $50,000 - 1000 = $49,000.
| Events | X | P(X) | x.P(x) |
| Lives | +1000 | 0.9999 | 999.90 |
| Dies | -49,000 | 1 – 0.9999 = 0.0001 | -4.90 |
| 1.0000 | 995.00 = |
The Expected value of a Discrete Random Variable
A life insurance policy that sells for $200 and should the person pass away before the end of the year, the family gets a check for $10,000 from the company.
The company can expect their profits divided by the number of policies sold (profit per policy) to be approximately $190.
The expected value is the long-run average after many, many trials, so while the company’s average profit is unlikely to ever be exactly $190, the more policies they sell, the closer and closer the company’s profit per policy will be to $190.
| Events | X | P(X) | x.P(x) |
| Dies | -9,800 | 0.001 | -9.80 |
| Lives | 200 | 0.999 | +199.80 |
| 1.0000 | $190.00 = |
Using Expected value to Distinguish between Two Possible Courses of Action
A bank can either risk $20,000 on a currency investment that has a 51% chance of earning them $40,000 in profit, or then can risk $700,000 on a bond investment that has a 98% chance of earning them $40,000. In the long run, which strategy will yield the most profit?
Currency: Bond:
Average profit from currency investment is $10,600 and that from the bond investment is $25,200. Thus the bond yields higher average profit, and is the better choice.
| x | P(x) | x.P(x) | |
| Profit | +40,000 | 0.51 | 20,400 |
| Loss | -20,000 | 0.49 | -9,800 |
| 1.00 | $10,600 = |
| x | P(x) | x.P(x) | |
| Profit | +40,000 | 0.98 | 39,200 |
| Loss | -700,000 | 0.02 | -14,000 |
| 1.00 | $25,2000 = |
The Variance and Standard Deviation of a Discrete Random Variable
The mean value for a discrete probability distribution provides the typical value for the random variable.
In other words, the mean tells us what we can expect to happen on average over the long run, but if we want to know how varied the outcomes for the random variable will be, we can calculate the variance or standard deviation for the random variable.
Variance for a probability distribution
The standard deviation for the random variable is found by taking t he square root of the variance of the random variable.
Standard Deviation for a probability distribution
The Variance and Standard Deviation of a Discrete Random Variable
Calculate the standard deviation of the probability distribution shown here (round to the thousandths place):
| x | P(x) | x.P(x) | ||
| 0 | 0 | 0.15 | 0 | 0 |
| 1 | 1 | 0.48 | 0.48 | 0.48 |
| 4 | 2 | 0.25 | 0.50 | 1.00 |
| 9 | 3 | 0.12 | 0.36 | 1.08 |
| 1.34 = | 2.56 = |
Determine if an Event is Unusual using the Mean and Standard Deviation of a Random Variable
A business venture offers an expected profit of $28,000 with a standard deviation of $5,250. Would it be unusual to earn less than $20,000 on the deal? (Hint: consider any value more than two standard deviations away from the mean as unusual)
Thus an earning of $20,000 is not unusual because it falls within the interval above.
4.3
The Binomial Probability Distribution
The Five Characteristics of a Binomial Experiment
A Binomial Experiment has a fixed number of trials, only two possible outcomes for each trial, one trial cannot affect outcome of the next trial, the probability has to remain constant from one trial to the next, and x must represent the number of successes.
Example: Flip a coin 3 times and count the number of heads that turn up (say, 1). Is this a binomial experiment?
There is a fixed number of trials, n = 3
There are 2 outcomes in each trial – heads and tails
The trials are independent, and the outcome of each flip does not affect that of the later flips
Each flip had a 50% chance of turning up heads
x = 1 here (success class is “heads”)
Binomial Probability Formula
The probability of having X successes out of n trials during a binomial experiment is given by the following formula (recall that :
where
n = the number of trials for the binomial experiment
x = the number o successes
p = the probability of a success
q = the probability of a failure (
Binomial Probability Formula
Example: If a binomial experiment involves slipping a fair coin 7 times and counting the number of heads that result, the probability of 5 heads turning up in 7 flips is provided below:
n = 7 (there are 7 flips of the fair coin)
x = 5 (we are looking for the probability of getting 5 heads)
p = 0.50 (a fair coin has a 50% chance of turning up heads on a single flip)
q = 0.50 (the probability of failure is found by subtracting the probability of success from 1)
Thus the probability of getting 5 heads out of 7 flips is 1.64%.
The Probability of X successes in a Binomial Experiment
A cable company believes that their new promotion will convince 20% of satellite television subscribers to sign up for cable. If the company is correct, what is the probability that 2 out of 8 randomly selected satellite users end up switching to cable after hearing the promotion?
The fact that there are 2 groups that will behave differently (some will switch and some will not) denotes it’s the binomial distribution.
Test to determine if this is a binomial experiment: (1) fixed number of trials = 8; (2) there are 2 possible outcomes (switches or not); (3) constant probability of success (switching) is 20%; (4) 8 unique users, so trials are independent (assume no user is called twice).
n = 8, X = 2 switch, p = 0.20, q = 1 – 0.20 = 0.80. So
Thus there is a 29.4% probability that 2 out of the 8 satellite users switch to cable after hearing the promotion.
The Probability of a Cumulative Set of Events for a Binomial Experiment
A drug company reports that 65% of balding men would benefit from using an over the counter hair-loss solution they manufacture. Assuming the company’s claim is correct and a random sample of 10 men are selected for a clinical trial of the product, what is the probability that at least 9 of the men benefit from the solution?
This is a binomial experiment because: (1) there is a fixed number of trials, 10 men selected; (2) each trial has 2 outcomes (working or not working); (3) constant probability of 65% of benefitting from the solution; (4) trials are independent (no chance of repetition)
Here n = 10, X = 9 or 10, p = 0.65, q = 1 – 0.65 = 0.35. So
P(X = 9 or X = 10) = P(X = 9) + P(X = 10)
= +
= 0.086
So there is an 8.6% chance that at least 9 men would benefit from the solution.
The Probability of a Cumulative Set of Events for a Binomial Experiment
A laptop manufacturer knows that 30% of its laptops will fail within the first two years of use. If seven randomly selected customers are surveyed, what is the probability that more than 3 of them experienced a laptop failure within the first two years of use?
This is a binomial experiment because: (1) there is a fixed number of trials – 7 men selected randomly; (2) there are two outcomes (a laptop fails or not within the first two years of use); (3) constant probability of 30% of a laptop failing in two years’ time; (4) customers are independent.
Here n= 7, X = more than 3 had a failure (4, 5, 6, or 7), p = 0.30, q = 1 – 0.30 = 0.70. So, from the table:
P(X=4) = 0.0972
P(X=5) = 0.0250
P(X=6) = 0.0036
P(X = 7) = 0.0002
Adding all these,
P(X more than 3) = 0.1260
Mean for a Binomial Probability Distribution
We could calculate the mean of the binomial probability distribution by listing all of the possible outcomes for the experiment, listing all of their corresponding probabilities, and then applying the formula:
However, that method could be very time consuming.
There is a simpler approach when trying to find the mean of a binomial probability distribution.
First we identify the number of trials for the experiment (n) and the probability of success (p). Then we apply the following formula:
Mean for a Binomial Probability Distribution
Example: If a quality control manager samples computer chips from a production line with replacement that have a 0.009 probability of being defective, what is the average number of defective chips that will be found in a sample of 200 chips?
This sampling procedure produces a binomial probability distribution, so we can apply the formula for finding the mean of a binomial distribution.
In this example, there are 200 trials, and the probability of finding a defective chip in 0.009
defective chips
For groups of 200 computer chips random selected from the production line, the average number of defective chips is 1.8.
The Mean of a Binomial Random Variable
An electronics retailer notes that only 8% of its online customers choose to purchase their extended service plan. If the retailer has 300 online sales over the next month, what is the expected number of customers that will choose to purchase the extended service plan?
This satisfies the conditions of a binomial experiment. So
since n = 300 and p = 0.08 here.
Thus the expected number of customers that will choose to purchase the extended service plan is 24.
Standard Deviation for a Binomial Probability Distribution
We could calculate the standard deviation for the binomial probability distribution by listing all of their corresponding probabilities, and then applying the formula
.
However, that method would be very time consuming.
There is a simpler approach when trying to find the standard deviation of a binomial probability distribution.
First, we identify the number of trials for the experiment (n) and the probability of success (p).
Then we apply the following formula:
Standard Deviation for a Binomial Probability Distribution
Example: If a quality control manager samples computer chips from a production line with replacement that have a 0.009 probability of being defective, what is the standard deviation for the number of defective chips that will be found in a sample of 200 chips?
This sampling procedure produces a binomial probability distribution, so we can apply the formula for finding the standard deviation of a binomial distribution.
In this example, there are 200 trials, and the probability of finding a defective chip in 0.009.
n = 200, p = 0.009, q = 1 – 0.009 = 0.991
Standard Deviation for a Binomial Probability Distribution
The variance will be
Then we simply take the square root to find the standard deviation:
defective chips
For groups of 200 computer chips randomly selected from this factory, the standard deviation for the number of defective chips is 1.336.
Standard Deviation of a Binomial Random Variable
A recent report states that only 28% of software projects were expected to finish on time and on budget. If we randomly sample 80 software projects, what is the standard deviation for the number of projects that are expected to finish on time and on budget?
This satisfies all the conditions for binomial experiment, so we can use the formula for the standard deviation to calculate this:
4.4
The Poisson Probability Distribution
The Poisson Distribution
The Poisson distribution is a discrete probability distribution that provides probabilities for the number of occurrences of some event over a given period, interval, distance, or space.
Example: A customer service call center might use the Poisson distribution to describe the behavior of incoming calls over different time periods.
Example: A website might use the Poisson distribution to estimate the likelihood of some number of individuals logging onto the site between the hours of 12:00AM and 1:00AM.
Example: A mining company might use the Poisson distribution to model the number of methane gas releases over a specified depth.
The Poisson distribution is typically used to model the occurrences of rare events.
The Poisson Distribution
The probability that the specified events occurs X times over some defined interval is given by the following formula:
where
(mu) is the mean (expected) number of occurrences (successes) over a particular interval
x is the number of occurrences (successes)
e is a constant (the base of the natural log) that is approximately equal to 2.71828
The Poisson Distribution
Here are some important characteristics of the Poisson distribution:
The random variable is the number of occurrences of some event over some defined interval.
The probability of the event is proportional to the size of the closed interval.
The intervals do not overlap.
The occurrences are independent of each other.
Mean and Standard Deviation of the Poisson Distribution
The mean of the Poisson distribution is
The standard deviation for the Poisson distribution is given by
Some Important Differences between the Binomial and the Poisson Distributions
The binomial distribution is dependent upon the sample size and the probability of success, while the Poisson distribution only depends on the mean .
In the binomial distribution, the random variable can take on values of 0,1,..,n, while in the Poisson distribution, the random variable can be any integer greater than or equal to zero.
In other words, there is no upper bound for the number of occurrences in the given interval.
Probability and the Poisson Distribution
A cellular communication company finds that during a 10-minute phone call there will typically be one incidence of poor reception. Use a Poisson distribution to calculate the probability that there will be 5 incidences of poor reception during a call that lasts an hour.
Here x = 5, = 6 since an average of 1 incidence of poor reception in a 10-minute period implies an average of 6 incidences in 60 mins (1 hour).
