Stats 151 - Mean, Standard Deviation, and Empire Rule for Random Variables and Binomial Distribution

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Chemistry 101 Stats 151 Flashcards on Stats 151 - Mean, Standard Deviation, and Empire Rule for Random Variables and Binomial Distribution, created by jennabarnes12387 on 28/01/2014.
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Question Answer
how do you find the mean of a random variable? sum of all x values times the probability = Ux = E(x)*p(x)
a gambler bets 10 dollars on red on a roulette wheel that have 18 red, 18 black, and 2 green slots. Whats his expected return? 10*(18/38) + (-10)*(20/38) = -0.526 on average the gambler will loose 52.6 cents per bet
how do you find the standard deviation of a random variable? sigma = square root(E(x-u)squared * p(x))
a gambler bets 10 dollars on a specific roulette number. he gets 350 if he wins with an average loss of 52.6 cents a bet. What is the standard deviation? sigma y = square root((350-(0.53)squared)*(1/38)+(-10 - (0.53))squared * 37/38 = 57.63
what is the probability you will fall into the middle section of a bell curve? 68% = 0.68 probability
What is a Bernoulli experiment? any experiment that has only two outcomes e.g. yes or no
how do you find the probability of the outcomes for a Bernoulli experiment p(outcome 1) = p p(outcome 2) = 1-p
what is the mean of a bernoulli experiment? the sigma? the mean is p. sigma is found by calculating (p(1-p))
what is a binomial experiment? a bernoulli experiment performed more then once
what do x and n-x equal in a binomial experiment? x = # of success' n-x = # of failures
How do you find the probability of x in a binomial experiment? (n over x) P to the x (1-p) to the n-x
What are the Bernoulli random variables represented as? How do they appear on a number line? the variables are presented as E1 and E2. E1 becomes 0 on the number line and E2 becomes 1
when you see n over x in brackets, what does it mean? nCx
Suppose that revenue Canada reports that 8% of all tax returns contain errors. An experiment consists of checking errors in a random sample of 10 tax returns. Find the probability that the sample contains exactly one error P(x=1) = (10 over 1)*0.8^(1)*(1-0.8)^(10-1) = 0.3777
at most one error? P(x< or equal to 1) = (x=0) + P(X=1) = (10 over 0) * 0.8^(0)(1-0.8)^(10-0) +0.3777 = 0.4344 + 0.3777 = 0.8121
at least one error? P(x>or equal to 1) = 1-P(x=0) = 1-0.4344 = 0.5656
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