Chapter
5 Discrete Random Variables and Their Probability Distributions
Mean and Standard Deviation of a Discrete Random Variable Computing the mean and standard deviation of a discrete random variable is slightly different than computing the mean and standard deviation of a set of data values. Each data value in the data set weighs equally in the computation. However, in a discrete random variable, the possible data values are given along with the likelihood of each value occurring on any given single trial. As was the case for a set of data values, the TI-84 calculator can be used to calculate the mean and standard deviation of a discrete random variable by either manually using the formulas or by using a built-in function. We will begin with manually using the formulas. Example: Number of Breakdowns Per Week Example 5-3 gives the probability distribution of the number of breakdowns per week for a machine based on past data. Enter the number of breakdowns into a list named X, and the probability into a list named PROBX.
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Mean of a Discrete Random Variable The formula to calculate the mean of a discrete probability distribution is . Move the cursor to highlight the name of the empty List next to List PROBX. Type: List X * List PROBX Press ENTER.
The formula now says to sum of this list of values. Go to the homescreen (2nd > Mode). Select 2nd > STAT > MATH > 5: sum( Type: L1) and press ENTER. µ = 1.8 breakdowns per week.
Standard Deviation of a Discrete Random Variable The formula to calculate the standard deviation of a discrete probability distribution is
Move the cursor to highlight the name of the empty List next to List PROBX. Type: List X ^ 2 * List PROBX Press ENTER.
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The formula now says to take the square root of the difference of the sum of this list of values and the square of the mean. Go to the homescreen (2nd > Mode). Press √ Key. Select 2nd > STAT > MATH > 5: sum( Type: L1) – 1.8 ^ 2) and press ENTER . σ = 1.03 breakdowns per week. Using TI-84 Plus Built-In Functions For Discrete Probability Distributions The TI-84 Plus built-in function 1-Var Stats will also calculate the numerical descriptive statistics for a Discrete Probability Distribution. We will use the same probability distribution as above, which we stored in Lists X and PROBX.
Select STAT > CALC > 1-Var Stats. Press ENTER.
Select 2nd > STAT >X , 2nd > STAT > PROBX Press ENTER. The screen will display the descriptive statistics, which includes the population mean and standard deviation.
Generating Dependent Probabilities Factorials A common function needed to compute dependent probabilities is the factorial function. The notation for the factorial of n is n! The “!” function is found on the MATH page under the PRB list. To find the number of ways six people could be arranged in six different chairs, you would calculate six factorial (6!).
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Type: 6 > MATH > PRB > 4: ! and press ENTER. 6! = 720. Calculate 10! and 0!.
Combinations The combination formula can also be used to compute dependent probabilities. The notation for the number of combinations is nCr, where n is the total number of elements, and r is the number being selected. Combinations are used when selecting a few elements from a larger number of distinct elements. Example: Ice Cream An ice cream parlor offers 6 flavors of ice cream. Kristen would like to purchase 2 flavors of ice cream. In how many ways can Kristen choose 2 flavors out of the 6 flavors? In order to find the number of ways of choosing two flavors out of six, we would need to calculate 6C2. Type 6 > MATH > PRB > 3: nCr and press ENTER. Press the number 2 key and press ENTER. There are 15 different combinations of two flavors of ice cream. Calculate 6C3 and 8C3 .
Permutations The permutation formula can be used to compute dependent probabilities. The notation for the number of permutations is nPr, where n is the total number of elements, and r is the number being selected. Permutations are used when trying to find all possible arrangements of elements taken from a larger selection. Arrangements involve putting the elements in a particular order.
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If Kristen’s story changes as below, then permutations apply rather than combinations. An ice cream parlor offers 6 flavors of ice cream. Kristen would like to purchase 2 flavors of ice cream and concerned as to which flavor is on the top and which flavor is on the bottom (i.e. Kristen is concerned about the arrangement of the flavors). In how many ways can Kristen arrange 2 flavors out of the 6 flavors? In order to find the number of arrangements of choosing two flavors out of six, we would need to calculate 6P2. Type 6 > MATH > PRB > 2: nPr and press ENTER. Press the number 2 key and press ENTER. There are 30 different arrangements of two flavors of ice cream. Calculate 6P3 and 8P3 .
Binomial Distribution Randomly Generating Number of Successes From a Binomial Distribution There are many situations in statistics where you need to generate numbers from distributions where the numbers are not equally likely to occur. One of the most commonly used distributions used in statistics is the discrete Binomial distribution. The TI-84 Plus has a built-in function to generate random real numbers from a specific Binomial distribution. The random real number represents an x value, the number of successes. Select MATH > PRB > 7:randBin( Type 3, 0.3,5) and press ENTER. The screen shot on the left repeated the Binomial experiment 5 times. Each time there were 3 trials with a probability of success were 2, 1, 0, 1, and 2 respectively. Generate 5 random numbers from a Binomial distribution with 3 trials and 0.9 probability of success.
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The syntax for the randBin( function is randBin(n, p, r). This will generate r random numbers representing x the number of successes from a binomial distribution with n number of trials and p probability of success on a given trial. Note: if r = 1, you may omit it.
Compute Binomial Probabilities The command for computing a probability at x successes for a discrete Binomial distribution is binompdf(. To find the probability of x successes out of n trials, each with probability p of success, type binompdf(n, p, x). Example: VCR’s Suppose that 5% of all VCR’s manufactured by an electronics company are defective. Three VCR’s are selected at random. What is the probability that exactly one of them is defective? P(x = 1) Select 2nd > VARS (DISTR) > A: binompdf( and press ENTER. Type: 3, 0.05, 1) and press ENTER. The result is 0.135375 or ≈ 13.5% chance that exactly one of them is defective. Calculate the same probability with 8 VCR’s selected at random, rather than 3. Now there is ≈ 27.9% chance that exactly one of them is defective.
Compute Cumulative Binomial Probabilities The command for the probability for a cumulative number of successes from 0 to x for a discrete Binomial distribution with n number of trials and p probability of success on any given single trial is binomcdf( . P(number of successes ≤ x) Using the same Binomial distribution of 3 VCR’s as above, what is the probability that zero or one of them is defective? P(x ≤ 1) Select 2nd > VARS (DISTR) > B: binomcdf( and press ENTER. Type: 3, 0.05, 1) and press ENTER.
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The result is 0.99275 or ≈ 99.3% chance that at most one of them is defective. Calculate the same probability with 8 VCR’s selected at random, rather than 3. Now there is ≈ 94.3% chance that at most one of them is defective. Compute Poisson Probabilities The command for computing the probability of x occurrences within a given interval for a discrete Poisson distribution with a mean number of occurrences λ is poissonpdf(λ, x). P(number of occurrences = x) Example: Telemarketing Suppose that a household receives, on average, 9.5 telemarketing calls per week. Find the probability that the household receives 6 calls this week. Select 2nd > VARS (DISTR) > C: poissonpdf( and press ENTER. Type: 9.5, 6) and press ENTER. The result is 0.076420796 ≈ 7.6% chance that the household receives 6 calls this week.
Find the probability that the household receives 10 calls this week. There is ≈ 12.4% chance that the household receives 10 calls this week.
Compute Cumulative Binomial Probabilities The command for computing the probability of at most x occurrences (cumulative) within a given interval for a discrete Poisson distribution with a mean number of occurrences λ is poissoncdf(λ, x). P(number of occurrences ≤ x) Using the same Poisson distribution of Telemarketing calls as above, what is the probability that the household receives at most 6 calls this week? P(x ≤ 6)
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Select 2nd > VARS (DISTR) > D: poissoncdf( and press ENTER. Type: 9.5, 6) and press ENTER. There is ≈ 16.5% chance that the household receives at most 6 calls this week.
Find the probability that the household receives at most 10 calls this week. There is ≈ 64.5% chance that the household receives at most 10 calls this week.
Geometric Probabilities Your calculator can compute probabilities for a geometric random variable with probability of success p using the geometpdf( command, located on the DISTR page. To find the probability of the random variable taking the value x, type geometpdf(p, x). Example: Car Ignition Suppose that a car with a bad starter can be started 90% of the time by turning on the ignition. What is the probability that it will take three tries to get the car started? Type geometpdf(0.9, 3); the answer is 0.9%. Cumulative Geometric Probabilities As with the binomial and cumulative probability functions, there is a cumulative version geometcdf( . It can be used to find the probability that a geometric random variable will take a value of at most x by typing geometcdf(p, x).
