Chapter 6: Random Variables

Chapter 6: Random Variables

If we tossed a coin 3 times...
How many tails could we possibly get? The possible # of tails (possible values of X) are 0, 1, 2, 3. Every time we toss a coin three times, X will either be 0, 1, 2, or 3 (and possibly different each time)

HHH HHT HTH THH HTT THT TTH TTT
What if… … we tossed a fair coin 3 times. With a partner, list all the possible outcomes HHH HHT HTH THH HTT THT TTH TTT This is the “sample space” for this chance process

Probability distribution…
8 equally likely outcomes; probability is 1/8 for each possible outcome. Define the variable X = the number of heads obtained. The value of X will vary from one set of tosses to another but will always be one of the numbers 0, 1, 2, or 3. How likely is X to take each of those values? It will be easier to answer this question if we group the possible outcomes by the number of heads obtained: Ask students, What’s the probability that: we get at least 2 heads? At most 1 head? Less than 1 head? Exactly 3 heads?

Random Variables ... X is called a random variable because its values vary when the coin tossing is repeated. are usually denoted by capital letters near the end of the alphabet, like X or Y

Discrete Random Variables
Discrete random variables have a “countable” number of possible positive outcomes and must satisfy two requirements. Note: ‘countable’ is not the same as finite. (1) Every probability is a # between 0 and 1; (2) The sum of the probabilities is 1 Two types of RV’s we will discuss: discrete and continuous; First we will discuss discrete. Note: countable is not the same as finite. Countable in the sense that we can list successive outcomes together with their associated probabilities. So, for example, the # of stars in the sky is discrete, even though, arguably, # of stars is infinite.

World-Wide 2015 AP Statistics Score Distribution
Is this a discrete random variable probability distribution? Why or why not? 1 2 3 4 5 .238 .189 .252 .132

Discrete Random Variables...
In your groups, discuss examples of discrete random variables Think-Pair-Share

Other Examples of Discrete Random Variables...
Number of times people have seen Fall Out Boy in concert Number of gifts we get on our birthday Number of burgers sold at In-N-Out Burger per day Number of stars in the sky All are whole, countable numbers; all vary; usually represented by a table or probability histogram

Non-examples of Discrete Random Variables...
Your height Weight of a candy bar Time it takes to run a mile These are continuous RV’s which will be discussed in a few; Also can discuss shoe sizes vs. foot length; shoes come in sizes (for example) 5, 5-1/2, 6, 6-1/2, etc. whereas foot length has an infinite number of possibilities. also time to run vs. # hurdles cleared; also age vs. # of birthdays you have had.

Mean of Discrete Random Variables
To find the mean (measure of center) of discrete random variable X, multiply each possible value by its probability, then add all the products Mean, Expected Value, μx Symbolic: Σxipi

Expected Value for Discrete Random Variables...
Grade 1 2 3 4 Probability 0.01 0.05 0.30 0.43 0.21 If 0 = F, 1 = D, etc., to calculate expected GPA for a given student, for example.

Find μx for the following discrete random variable: Mean =

Find μx for the following discrete random variable: Mean = 1.75

Standard Deviation for Discrete Random Variables...
Standard deviation measures the spread or variability of the distribution (remember mean, SD, variance; median, IQR) Symbol: σ (σ)2 = variance; σ 2 = σ Don’t recommend using formula to calculate variance and SD; use calculator & lists

Standard Deviation for Discrete Random Variables...
L1 random variables X L2 corresponding probabilities stat – calc – 1-var stats L1, L2 (L1, L2 is only way calculator knows you have a frequency distribution) 𝑥 is really μx (or expected value); σx is SD

Standard Deviation & Expected Value of Discrete Random Variables...
μx = expected value = σx = standard deviation = σ2 = variance = Cars Sold 1 2 3 Probabilities 0.3 0.4 0.2 0.1

Standard Deviation & Expected Value of Discrete Random Variables...
μx = expected value = 1.1 σx = standard deviation = σ2 = variance = 0.890 Cars Sold 1 2 3 Probabilities 0.3 0.4 0.2 0.1

Go back to grade distribution
Use calculator to confirm the expected value; and calculate the standard deviation and variance using your calculator μx = σ = σ 2 = Grades 1 2 3 4 Probability 0.01 0.05 0.30 0.43 0.21

Go back to grade distribution
Use calculator to confirm the expected value; and calculate the standard deviation and variance using your calculator μx = σ = σ 2 = .752 Grades 1 2 3 4 Probability 0.01 0.05 0.30 0.43 0.21

Continuous Random Variables ...
take on all values in an interval of numbers probability distribution is described by a density curve probability of event is area under the density curve and above the values of X that make up the event Bring back shoe size vs. length of foot; time vs. # of hurdles cleared; age vs. # birthdays

Continuous Random Variables...
are usually measurements heights, weights, time amount of sugar in a granny smith apple, time to finish the New York marathon, height of Mt. Whitney Are we ‘sure’ that the given runner’s time was 4 hours and 36 minutes? We know the exact time only as accurately as the rounding we choose to use. Discuss height in this manner (for Mt. Whitney; by the way, Mt. Whitney is said to be 14, 494 feet tall), discuss grams of sugar in a granny smith apple.

How can we distinguish between continuous and discrete?
Discuss in your groups for a few minutes. Ask yourself ‘How many? How much? Are you sure?’ For example, # of children, pounds of Captain Crunch produced each year, # of skittles, ounces in a bag of skittles TPS; Pounds of CC vs. # of cereal bits of CC; remind students about discrete does not necessarily mean finite.

Probability distribution is area under the density curve, within an interval, above x-axis Ask students if this looks familiar? Like Normal distributions…

For continuous RV’s, there is no difference between > & and no difference between < & Ask students why they think this is the case; will get back to explanation in a few minutes.

Continuous Random Variables ...
This is true for all continuous random variables, but NOT discrete random variables. Why? Discuss for a few minutes. 𝑃 1<𝑋<4 is the same as 𝑃 1≤𝑋≤4 where x = # of Reeses Peanut Butter Cups I will eat in a given day (yum yum!)

Random Variables... Consider a six-sided die... What is the probability... P ( roll less than or equal to a 2) = P ( roll less than a 2) = Different probabilities; discrete random variable Note: possible outcomes are 1, 2, 3, 4, 5, 6; but probabilities of those outcomes are (often) fractions/decimals

Continuous random variables ...
All continuous random variables assign probabilities to intervals All continuous random variables assign a probability of zero to every individual outcome. Why?

There is no area under a vertical line (sketch) Consider... to P = 0.02 to P = 0.002 to P = P (an exact value –vs. an interval--) = 0

Another way to think about this is …
For continuous RV’s, there is no area (so probability = 0) at a given point… Remember for continuous RV’s each outcome is just one of an infinite number of possible outcomes, so the probability of any particular outcome is 1 ∞ ► 0 As the value of n increases, the closer 1 𝑛 gets to zero

Density Curves & Continuous RV’s..
Can use ANY density curve to assign probabilities/model continuous RV’s; many models Most familiar density curves are the Normal density curves (Chapter 2)

Remember Normal density curve... N ( ) N (0, 1) 2nd – stat- normal cdf Always label ‘low, high, Mu, Sigma’ for full credit. Calculator speak not sufficient. Label input and sketch a density curve.

Mean & Standard Deviation of Continuous Random Variables...
μx for continuous random variables lies at the center of a symmetrical (or fairly symmetrical) density curve (Normal or approximately Normal, Chapter 2) N (μ, σ) Calculating σ and/or σ2 for continuous random variables…. beyond the scope of this course… will be given this information if needed

Practice…. … with discrete and continuous random variables and general probability Now lets do some general practice for the next few problems….

Cheating in school... A sample survey asked an SRS of 400 undergraduates: “You witness two students cheating on a quiz. Do you go to the professor?” Twelve percent answered ‘yes.’ This statistic is a random variable, because repeating the SRS would give a different sample of 400 undergraduates and a different proportion (other than 12%). This distribution of a sample proportion is N(0.12, 0.016). What is the probability that a given survey result differs from the truth about the population by more than 2 percentage points? Now lets do some general practice for the next few problems….

Cheating in school... About 21% of sample results will be off by more than two percentage points. Continuous RV; students must show all work.

Car Ownership Choose an American household at random and let the random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: Ask students if this is discrete or continuous. Why?

Verify that this is a legitimate discrete distribution.
(b) Say in words what the event {X ≥ 1} is; then find P(X ≥ 1). (c) A housing company builds houses with two-car garages. What percent of households have more cars than the garage can hold? Also have students explain why > 1 is different from part (b).

How Student Fees Are Used...
Weary of the low turnout in student elections, a college administration decides to choose an SRS of 3 students to form an advisory board that represents student opinion. Suppose that 40% of all students oppose the use of student fees to fund student interest groups and that the opinions of the 3 students on the board are independent. Then the probability is 0.4 that each opposes the funding of interest groups. (a) Call the three students A, B, and C. What is the probability that A and B support funding and C opposes it?

How Student Fees Are Used ...
(a) Call the three students A, B, and C. What is the probability that A and B support funding and C opposes it? P (Student A Supports & Student B Supports & Student C opposes) P (SSO) = (0.60) (0.60) (0.40) = 0.144

(b) List all possible combinations of opinions that can be held by Students A, B, and C. (Hint: There are eight possibilities.) Then give the probability of each of these outcomes. Remember... Support = Oppose = 0.40

SSS, SSO, SOS, OSS, SOO, OSO, OOS & OOO SSS SSO SOS OSS SOO OSO OOS OOO (.6)3 (.6)2(.4) (.4)2(.6) (.4)3

(c) Let the random variable X be the number of student representatives who oppose the funding of interest groups. Give the probability distribution of X.

(c) Let the random variable X be the number of student representatives who oppose the funding of interest groups. Give the probability distribution of X. P(X=0) P(X=1) P(X=2) P(X=3) 0.216 0.432 0.288 0.064

(d) Express the event “a majority of the advisory board opposes funding” in terms of X and find its probability.

(d) Express the event “a majority of the advisory board opposes funding” in terms of X and find its probability. This probability can be written two different ways: P (X > 1) P (X ≥ 2) Either equal to 0.352 Follow up: ask/class discussion, in general, not necessarily this particular slide/situation: P (x > 1) vs. P (x >= 2); same or different? Depends on if we are talking about discrete RV or continuous RV.

Homework... Page 359, #1, 3, 5, 7, 9, 11, 13, 15, 17, 21, 23, 25 MC: all FRQs: as needed

Transforming & Combining RV’s (Changing Units of Measure)
Remember Chapter 2…the last time you went out to dinner. How much was the bill? What is 𝑥 ? What is s? How about other key values, like Q1, Q3, etc.? Tipped chef $10; what did that do to our mean, standard deviation, spread, shape of distribution? Paid 20% more; what did this do to our mean, standard deviation, spread, shape of distribution? No matter if they paid or if someone else paid; if they don’t know how much, have them make a good guess.

Linear Transformation
A linear transformation changes the original variable x into the new variable xnew given by an equation of the form xnew = a + bx Adding/subtracting the constant a shifts all values of x upward (increases; to the right) or downward (decreases; to left) by the same amount. Multiplying/dividing by the positive constant b changes the size of the unit of measurement.

Linear Transformation Effects
Adding the same number a (either positive, zero, or negative) to each observation adds a to measures of center and to quartiles but does not change measures of spread or shape. Multiplying/dividing each observation by a positive number b multiplies both measures of center (mean and median) and measures of spread (interquartile range and standard deviation) by b but does not change shape.

If you are a formula person… (Linear Transformations)
If X and Y are random variables, and a & b are fixed (constant) numbers, then μ a + bx = a + bμx σ2a+bX = b2σ2X

RV Means, SDs, & Variances: Example...
In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. The temperature of the flame varies a bit. Below is the distribution of the temperature X measured in degrees Celsius:

RV Means & SD/Variances
The mean temperature μx = 550 degrees Celsius & standard deviation, σx = 5.7 degrees Celsius A manager asks for results in degrees Fahrenheit. The conversion of X into degrees Fahrenheit is y = 9/5 X Find the mean μx & standard deviation σx in Fahrenheit μ a + bx = a + bμx σ2a+bX = b2σ2X Careful to emphasize that we perform the linear transformation on the variable (temp) only, not the probability; applying the LT to prob would not make any sense; discuss.

RV Means & SD/Variance The conversion of X into degrees Fahrenheit is y = 9/5 X Find the mean μx in Fahrenheit. Mean = 1022 degrees Fahrenheit SD = degrees F.

RV Means, etc. Can choose to use the formula or...
L3= (L1) (( 9/5) X + 32), then do 1-var stats with L3 and L2 Let’s try it

You try it... Hart Partners is planning a major investment. The amount of profit X is uncertain, but an estimate gives the following distribution (in millions of dollars): Find the mean profit μx and the standard deviation σx Hart 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 from the investment. Find the mean and the standard deviation of what it expects to actually retain

You try it... Hart Partners is planning a major investment. The amount of profit X is uncertain, but an estimate gives the following distribution (in millions of dollars): Find the mean profit μx = $3 million; standard deviation $2.52 million The firm actually retains Y = 0.9X -0.2 from the investment. Find the mean of what it expects to actually retain = $2.5 million; SD = $2.27 million

RVs: Combining Means & Standard Deviations/Variances
If X and Y are random variables, then μX+Y = μX + μY … remember, means play nice The mean of the sum of the random variables is the sum of their means You earn an average of $20/week allowance μX = $20 Your sister earns on average $16/week allowance μY = $16 So, μX + μY = $36

RVs: Combining Means & Standard Deviations/Variances
If X and Y are independent random variables, then σ2X+Y = σ2X + σ2Y σ2X-Y = σ2X + σ2Y … variances? They don’t play so nice Notes: (1) Go through variances always; (2) always add variances; (3) constant “a” doesn’t change/ influence standard deviation/variance

Combining Normal Random Variables
Any linear combination of independent Normal random variables is also going to be Normally distributed If X and Y are Normal independent random variables, and if a and b are fixed numbers, then aX plus ± minus bY is also Normally distributed

Gold Medalist Carly Patterson
Carly Patterson won the gold medal in gymnastics at the 2004 Olympics in Athens, Greece. A competitor's total score is determined by adding the scores for four events: vault, parallel bars, balance beam, and floor exercise. Suppose we know that the eventual silver medalist, Russia's Svetlana Khorkina, earned a total score of (her actual total in Athens). Suppose also that Carly's scores in 100 previous meets leading up to the Olympics have been approximately Normally distributed.

We will further assume that Carly's perfomance at the Olympics will follow the same pattern as her pre-Olympic meets. This is a reasonable assumption for world-class athletes like Carly. We want to know Carly's chances of beating the total score of the current leader, Khorkina.

μv + μp + μb + μf = σ2v + σ2p + σ2b + σ2f = so σv+p+b+f =

μv + μp + μb + μf = points σ2v + σ2p + σ2b + σ2f = , so σv+p+b+f = Carly’s total score has a distribution that can be described as …

μv + μp + μb + μf = points σ2v + σ2p + σ2b + σ2f = , so σv+p+b+f = Carly’s total score will be N (37.871, ) P (Carly’s total > Khorkina) =

μv + μp + μb + μf = points σ2v + σ2p + σ2b + σ2f = , so σv+p+b+f = Carly’s total score will be N (37.871, ) P (Carly’s Score > ) =

Speed Dating… To save time and money, many single people have decided to try speed dating. At a speed dating event, women sit in a circle and men spend about 10 minutes getting to know a woman before moving on to the next one. Suppose that the height M of male speed daters follows a Normal distribution with a mean of 69.5 inches and a standard deviation of 4 inches, and that the height F of female speed daters follows a Normal distribution with a mean of 65 inches and a standard deviation of 3 inches. Alternative example (page 381 of text of TE shows all steps)

Speed dating… What is the probability that a randomly selected male speed dater is taller than the randomly selected female speed dater with whom he is paired? There is about an 82% chance that a randomly selected male speed dater will be taller than the randomly selected female speed dater with whom he is paired. Alternative example (page 381 of text of TE shows all steps)

Homework Page 382, # 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 61 MC FRQs

Binomial & Geometric Random Variables

Many types of distributions...
What is the main type of distribution that we have worked with (and is really valuable to us) so far? Normal distribution Characteristics of Normal distribution? (fairly) Symmetric, uni-modal, , area of one (all density curves), NPP is (fairly) linear.

Many other important, “special” types of distributions...
If certain criteria is met, easier to calculate probabilities in specific situations Next types of distributions we will examine are situations where there are only two outcomes Win or lose; make a basket or not; boy or girl ...

Discuss situations where there are only two outcomes...
Yes or no Open or closed Patient has a disease or doesn’t Something is alive or dead Person has a job or doesn’t A part is defective or not TPS

that is what this section is all about...
... two classes of distributions that are concerned about events that can only have 2 outcomes Binomial Distribution & Geometric Distribution

Binary; Independent; fixed Number; probability of Successes
The Binomial setting is: Each observation is either a success or a failure (i.e., it’s binary) All n observations are independent Fixed # (n) of observations Probability of success, p, is the same for each observation “BINS” Die; even or odd; 2 rolls; prob same; independent rolls; check.

Binomial Distribution: practice…
I roll a die 3 times and observe each roll to see if it is even or odd. Is: each observation is either a success or a failure? all n observations are independent? fixed # (n) of observations? probability of success, p, is the same for each observation ? BINS Die; even or odd; 2 rolls; prob same; independent rolls; check.

Binomial Distribution
If BINS is satisfied, then the distribution can be described as B (n, p) B binomial n the fixed number of observations p probability of success Note: This is a discrete probability distribution. Remember N (μ, σ)… is this discrete??

Binomial Distribution
Most important: being able to recognize situations and then use appropriate tools for that situation Let’s practice...

Are these binomial distributions? Why or why not?
Toss a coin 20 times to see how many tails occur. Asking 200 people if they watch ABC news Rolling a die until a 6 appears Yes, yes, no

Are these binomial distributions or not? Why or why not?
Asking 20 people how old they are Drawing 5 cards from a deck for a poker hand Rolling a die until a 5 appears No, no, no

How could we change the situations to make these/force these to be binomial distributions?
Asking 20 people how old they are Drawing 5 cards from a deck for a poker hand Rolling a die until a 5 appears

Binomial Distribution... Is the situation ‘independent enough’?
An engineer chooses a SRS of 20 switches from a shipment of 10,000 switches. Suppose (unknown to the engineer) 12% of switches in the shipment are bad. Not quite a binomial setting. Why? For practical purposes, this behaves like a binomial setting; ‘close enough’ to independence; as long as sample size is small compared to population. Rule of thumb: sample ≤ 10% of population size

Binomial Formula. you don’t need to memorize
Binomial Formula...you don’t need to memorize... Just know what ‘n’ (# of observations) and ‘k’ (value of random variable) Won’t ever use this; will use binomcdf or pdf

An engineer chooses an SRS of 20 switches from a shipment of 10,000 switches. Suppose that (unknown to the engineer) 12% of switches in the shipment are bad. The engineer wants to find the probability of exactly 1 switch failing. Situation ≈ Binomial k = 1 n= 20 p = 0.12

Binomial probabilities: PDF
2nd-vars binompdf (trials, p, X value) Use for exact value (i.e., probability that a family has exactly 2 boys) Remember, these are discrete random variables Can calculate probabilities of exact values (unlike continuous random variables) P, pointing, particular

Binomial probabilities: CDF
2nd-vars binomcdf (trials, p, X value) Use for cumulative values Works like table A; cumulative to left If want area to right, need to do “1 minus ....” Pay attention to < vs. ≤; and > vs. ≥

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that exactly 2 of them have type O blood? Binomial setting? Check for BINS. p = n = 5 X = 2 binompdf (n, p, X) = binompdf (5, .25, 2) = = ; context, always! Not sure if it will be child #1 & #2 or #1 & #5 or #4 & #3....etc. Just exactly 2 but not sure which 2

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that exactly 4 of them have type O blood? Binomial setting? Check for BINS. p = n = 5 X = 4 binompdf (n, p, X) = binompdf (5, .25, 4) = = ; context, always

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that exactly 1 of them have type O blood? Binomial setting? Check for BINS. p = n = 5 X = 1 binompdf (n, p, X) = binompdf (5, .25, 1) = = ; context, always

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that at most 2 of them have type O blood? Binomial setting? Check for BINS. p = n = 5 X = 2 binomcdf (n, p, X) = binomcdf (5, .25, 2) = = ; context, always

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that at most 4 of them have type O blood? Binomial setting? Check for BINS. p = n = 5 X = 4 binomcdf (n, p, X) = binomcdf (5, .25, 4) = = , context, always

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that at most 1 of them have type O blood? Binomial setting? Check for BINS. p = n = 5 X = 1 binomcdf (n, p, X) = binomcdf (5, .25, 1) = = , context, always

Practice... ... probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that at least 2 (meaning 2, 3, 4, or 5) of them have type O blood? p = n = 5 X = 2, 3, 4, or 5 1 – (binomcdf (5, .25, 1)) = = ; context, always

Practice... p = 0.25 n = 5 X = 3, 4, or 5 1 – (binomcdf (5, .25, 2)) =
... probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that at least 3 (meaning 3, 4, or 5) of them have type O blood? p = n = 5 X = 3, 4, or 5 1 – (binomcdf (5, .25, 2)) = = ; context, always

Practice... p = 0.25 n = 5 X = 4 or 5 1 – (binomcdf (5, .25, 3)) =
... probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that more than 3 (meaning 4 or or 5) of them have type O blood? p = n = 5 X = 4 or 5 1 – (binomcdf (5, .25, 3)) = = ; context, always

Practice... p = 0.25 n = 5 a) …at most 3 of them…
... probability 0.25 of having blood type O. If these parents have 5 children, what is the probability that … of them have type O blood? p = n = 5 a) …at most 3 of them… b) …at least 4 of them… c) …more than 1 of them… d) …exceeds 3 of them… e) … below 2 of them… f) … 0 of them…

Practice... Suppose the mean number of children in a given household who have type O blood is 2.1 with a standard deviation of If this distribution is approximately Normally distributed, what is the probability that a household has more than 4 children with type O blood? μ= 2.1 σ= 0.8 normalcdf (4, hi, 2.1, 0.8) = ; context, always

Caution... binomcdf (n, p, X). What is P(X) if...
Suppose X is B (23, 0.7). . . P (X < 17) = binomcdf (23, 0.7, 16) P (X ≤ 17) = binomcdf (23, 0.7, 17) Helpful hint... make a list of values you are interested in...

Caution... binomcdf (n, p, X). What is P(X) if...
Supposed X is B (23, 0.7). . . P (X > 21) = 1 - binomcdf (23, 0.7, 21) interested in 22, 23 P (X ≥ 21) = 1 - binomcdf (23, 0.7, 20) interested in 21, 22, 23

Binomial Distribution: Mean and Standard Deviation
If a basketball player makes 75% (“p”) of her free throws, what do you think the mean number of baskets made will be in 12 tries? Discuss. (0.75) ( 12) = 9; we expect she should make 9 baskets in 12 tries We expect her 𝑥 = 9 baskets

Binomial Mean & Standard Deviation
If a count X is a binomial distribution with number of observations n and probability of success p, then μ= np and σ= Only for use with binomial distributions; remember criteria... BINS Emphasize we don’t want to use these formulas with any RVs… they must be binomial RVs

Practice... If a basketball player makes 75% (“p”) of her free throws, we expect her to make 9 baskets in 12 tries. What is the SD of this distribution? SD = = = 1.5

Practice... Each child born to a particular set of parents has probability 0.25 of having blood type O. If these parents have 5 children, what is the mean and standard deviation for this distribution? mean = np = (5)(.25) = 1.25; the family should expect to have 1.25 children with type O blood SD = = = Emphasize that binomials distributions are count data; can’t be fractions/decimals; but MEANS of discrete RV (binomials) CAN be fractions/decimals.

Remember & Caution... Binomial distribution is a special case of a probability distribution for a discrete random variable All binomials distributions are discrete random variable distributions BUT not all discrete random variable distributions are binomial distributions Don’t assume; don’t apply binomial tools to all discrete RV distributions; must meet binomial distribution criteria (BINS)

Normal Approximation to Binomial Distribution...
But as ‘n’ gets larger, binomial distribution ≈ Normal distribution (weird... binomial is discrete & Normal is continuous...) General rule: np ≥ 10 & n(1-p) ≥ 10, the binomial distribution is ≈ Normal, i.e., the expected number of successes and failures are both at least 10; ‘large count condition’

Normal ≈ to Binomial Distribution...
Attitudes towards shopping... SRS of 2,500 (out of 218 million U.S. adults) asked if they found shopping frustrating or not Suppose in fact 60% of all U.S. adults say shopping is frustrating. Is this a binomial distribution? Check BINS.

Binomial, Simulation, or Normal Distribution...
BINS: 2 (frustrated or not), i (independent enough), n = 2,500 p = 0.60; confirmed binomial Say we want to find P(X ≥ 1520), we could: use binomcdf do a simulation normalcdf

P(X ≥ 1520), Using Binomial Distribution...
binomcdf: P( X ≥ 1520) = P( X ≥ 1520) = 1 – binomcdf (2500, 0.60, 1519) = 1 – = ; context, always Why is our x value 1519? Discuss.

P(X ≥ 1520), Using Simulation… n = 2500; p = 0.60
How would you suggest using a simulation to model this situation and calculate a probability? Discuss for 2 minutes, then share out TPS

P(X ≥ 1520), Using Normal ≈ to Binomial Distribution...
The binomial distribution ≈ N (μ, σ) due to np ≥ 10 & n(1-p) ≥ 10 (expected values for success & failures both ≥ 10) Calculate μ and σ μ = σ = 24.49

P(X ≥ 1520), Using Normal ≈ to Binomial Distribution...
Binomial ≈ N (μ, σ) = N(1500, 24.49) P (X ≥ 1520) = P (X > 1520) (in Normal setting) normalcdf (1520, hi, 1500, 24.49) = ; context, always (can also use table A... but lots of work) Remind students to label input for normcdf, low, high, mu, sigma

Histogram: B(2500, 0.60) Density Curve: N(np, √np(1-p))= N(1500, 24.49)

binomcdf = normalcdf = Table A, simulation – different but similar All are acceptable answers; clearly explain your process, justify work, answer in context, etc.

B (n, p)  n large  ≈ Normal (np, √np(1-p)) Rule of thumb: can use Normal approximation when np ≥ 10 & n(1-p) ≥ 10 Accuracy of Normal ≈ improves as n increases Most accurate for any fixed n when p ≈ 0.50 Least accurate when p is near 0 or 1 (skewed distribution)

The Geometric Distribution...
Remember binomial distribution BINS Geometric: BI_S; no ‘n;’ no fixed number of trials In Geometric setting, it is the number of trials required to obtain first success

Geometric Distribution...
geometpdf (p, X) (exactly) X = first success will happen on ___ trial geometcdf (p, X) (cumulatively) X ≤ first success will happen at most on ___ trial

Geometric Distribution ...
Rolling a die until a 5 appears BI_S? it either is a 5 or not rolls are independent no fixed n p = 1/6

What is the probability we roll exactly 3 times to get our first 5? P(X = 3) = geometpdf (1/6, 3) = , context What is the probability we roll exactly 100 times to get our first 5? P(X = 100) = geometpdf (1/6, 100) ≈ 0, context What is the probability we roll exactly 6 times to get our first 5? P(X = 6) = geometpdf (1/6, 6) = , context

What is the probability it takes us up to 3 times to get our first 5? P(X ≤ 3) = geometcdf (1/6, 3) = , context ... up to 100 times.... P (X ≤ 100) = geometcdf (1/6, 100) = , context ... up to 6 times.... P (X ≤ 6) = geometcdf (1/6, 6) = , context

What is the probability it takes us more than 3 rolls to get our first 5? P(X > 3) = 1 - geometcdf (1/6, 3) = = , context ... more than100 times.... P (X > 100) = 1 - geometcdf (1/6, 100) = = ≈ 0, context ... more than 6 times.... P (X > 6) = 1 - geometcdf (1/6, 6) = = , context

Mean & SD for Geometric Distributions...
Mean = Expected Value = μX = E (X)= 1/p Example: What is the theoretical probability of rolling a 2? 1/6 ≈ E(X) = 1/p = 1/( ) = 6; makes sense; we would expect it to take (on average, in the long run) six rolls of a die to roll a 2.

Mean & SD for Geometric Distributions...
Standard deviation = What is the standard deviation of this distribution? SD = = =

Homework Remember, always use technology (your calculator) to calculate probabilities Page 410, #69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 95, 97 MC, FRQs, FRAPPY, Chapter Review, AP Practice Test, etc. Good activities to include are Casino games, potato chips, and is this your lucky day (from publisher’s resources)

Chapter 6: Random Variables

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