binomial distributionbinomial settingnumber x of switchesbinomial probabilitieswhatindividual timesnumber x of red cardsindependent observationsblood types

There are 4 runners on the New High School team. The team is planning to participate in a race in which each runner runs a mile. The team time is the sum of the individual times for the 4 runners. Assume that the individual times of the 4 runners are all independent of each other. The individual times, in minutes, of the runners in similar races are approximately normally distributed with the following means and standard deviations.

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Runner 3 thinks that he can run a mile in less than 4.2 minutes in the next race. Is this likely to happen? Explain.The distribution of possible team times is approximately normal. What are the mean and standard deviation of this distribution?Suppose the teams best time to date is 18.4 minutes. What is the probability that the team will beat its own best time in the next race?

2NIP2 outcomes: each observation can be categorized as a success or failureN: Fixed # of observations I: Observations are independentP: The probability is the same for each observationExamples:Coin toss to see which of the 2 football teams gets the choice of kicking off or receiving to begin the game A couple prepares for their first child

EX 8.1If both parents carry genes for the O and A blood types, each child has probability = .25 of getting two 0 genes resulting in blood type 0. Assume independence.A success is the number of 0 blood types among 5 children of these parents in 5 independent observations.Is this a binomial distribution?

Ex 8.2You are dealt 10 cards from a shuffled deck and wish to count the number X of red cards as you go through the deck.A success is a red card.Is this a binomial distribution?

How to calculate Binomial Probabilities

What this means: Although sometimes a problem is not strictly independent and therefore not a binomial setting, the situation is close enough for practical purposes.

Example 8.3An engineer selects an SRS of 10 switches from a large shipment for detailed inspection. Unknown to the engineer, 10% of them fail to meet the specifications. What is the probability that no more than 1 of the 10 switches in the sample fail inspection?

(Note: Assuming that a defective switch is drawn first (p=0.1), the probability for the second switch being defective changes to p=0.0999. For practical purposes, this behaves like a binomial setting even if the condition of independence does not strictly hold).

Ex. 8.6: The number X of switches that fail inspection in Example 8.3 has approximately the binomial distribution with n = 10 and p = .1. Find the probability that no more than 1 switch fails.

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CalculatorWe want to calculate when X is B(10, .1)P(X 1) = P(X=0)+P(X=1)TI-83: the function binompdf(n,p,x) calculates thisP(X=0): Binompdf(10,.1,0) = .3486784401P(X=1): Binompdf(10, .1, 1) = .387420489P(X=0)+P(X=1) = .3487 + .3874 = .7361

PDFGiven a discrete random variable X, the probability distribution function assigns a probability to each value of X. The probabilities must satisfy the rules for probabilities given in chapter 6. B(5, .25) distribution where n=5 children

PDF by calculatorHow we arrived at the probability distribution function for B(5, .25) distribution where n=5 childrenBinompdf(5, .25, 0), Binompdf(5, .25, 1).Binompdf(5, .25,5)

Alternative Calculation/shortcut! CDFGiven a random variable X, the cumulative distribution function of X calculates the sum of the probabilities for 0, 1, 2up to the value of X.I.e., it calculates the probability of obtaining at most X successes in n trials. For last example: P(X 1) = binomcdf(10, .1, 1) = .736098903 (same answer, less work)

Example 8.8Corinne is a basketball player who makes 75% of her free throws. In one game, she shoots 12 free throws and makes only 7 of them. Find the probability of making a basket on at most 7 free throws.Check Binomial Settings (Use binompdf(n,p,X))P(X 7) = P(X=0)+P(X=1)+P(X=2).+P(X=7)= .0000+.0000+.0004+.0024+.0115+.0401+.1032= .1576Same answer, less work: P(X 7) = binomcdf(12, .75, 7) = .1576436761