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Statistical Distributions Example 1: Solution: Example 3: Solution: Example 4: Solution: , so therefore Expected Value of a discrete random variable X Expected value is the same as finding the mean. We can do this by using a modification of the mean equation, which looks like this: Example 1: Solution using the equation: We can also use technology to solve this: Ti-84 – Put x under L1 and P(X=x) under L2 under STAT. After this is entered go to STAT->CALC and select 1-Var Stats then enter L1, L2 after 1-Var Stats then ENTER. Example 2: Solution: 1 0.1 2 0.3 3 0.6 (You can use your calculator to find the standard deviation). Example 3: Solution: 1 1/36 2 3/36 3 5/36 4 7/36 5 9/36 6 11/36 Example 4: Solution: a) b) c) Since you spend $2 on a ticket and expect to win $1.7 you actually lose $0.3 per ticket on average. Binomial Distribution We often write a binomial distribution as If we go back to Example 1 we can use this method since this is a binomial distribution. The number of trials is 3 (n = 3) and the probability of success is getting a six (p = 1/6). So if we want to find the probability of getting 2 sixes we can use the following setup: We can also use the Ti-84 calculator going under 2nd VARS (DISTR)->binompdf . It is set up like this binompdf(n,p,r) so for our question it will be binompdf(3,1/6,2) = 0.0694. On the nspire go to menu->Probability->Distributions->Binomial Pdf Note as well that Binomial Cdf represents the cumulative probability for . So if then . On the calculator we would enter Ti-84: binomcdf(3,1/6,2) or in Nspire it would be binomCdf(3,1/6,0,2). Example 1: a) P(X=6)=binompdf(6,0.5,6)=0.0156 b) P(X=2)=binompdf(6,0.5,2)=0.234 c) P(X d) P(X )=binomcdf(6,0.5,1)=0.109 )=1 - P(X =0.344 )=1-binomcdf(6,0.5,3) Example 2: a) P(X=7)=binompdf(7,0.6,7)=0.0280 b) c) P(X=6)=binompdf(7,0.6,6)=0.131 d) P(X )=1 - P(X =0.710 )=1-binomcdf(7,0.6,3) For binomial distributions where n is the number of trials and p is the probability of success: Mean of x is E(X)= Variance of x is Var(X)= and therefore the standard deviation is Example 8: a) 0 0.0156 b) c) 1 0.0938 2 0.2344 3 0.3125 4 0.2344 5 0.0938 6 0.0156 Example 9: Solution: If the random variable X is normally distributed it can be written as . In this diagram there are 3 normal distributions with the same standard deviation but not the same mean. In this diagram there are 3 normal distributions with the same mean but not the same standard deviation. Example 4: a) b) c) = 0.3413 + 0.1359 =0.4772 = 0.3413 + 0.1359 + 0.0215 =0.4987 = 0.9544 + 0.0215 =0.9759 In statistics we use a standard normal distribution where the mean is 0 and a standard deviation is 1. We use Z as the random variable so under a standard normal curve . We can use technology to find the area (probability) . Using your GDC: Ti-84: 2nd VARS -> normalcdf(Lower Bound, Upper Bound) Nspire: menu->Statistics(6)->Distributions(5)->Normal Cdf (Lower Bound, Upper Bound) Example 1: Given a) b) find the following: c) d) Solutions: a) 0.5 b) 0.159 or 1 – (0.5 + 0.3413) c) 0.819 d) 0.961 Since most distributions do not match the shape of the standard normal distribution we can convert it using the equation: , if ) To solve most of these types of questions do the following: (IB often requires this as work) 1. Draw a graph of the distribution. 2. Convert x to z using the equation. You can also use your GDC: Ti-84: 2nd VARS -> normalcdf(Lower Bound, Upper Bound, mean, s.d.) Nspire: menu->Statistics(6)->Distributions(5)->Normal Cdf (Lower Bound, Upper Bound, mean, s.d) So if you wanted to find the probability that X was between 1.3 and 4.5 if ) then you could enter in your calculator normalcdf(1.3,4.5,1.4,2.24) or 0.435. (note that you should only use this if you do not need to show work) Example 2: If Z has a standard normal distribution find a: a) b) Solutions: (make a sketch for each) a) invNorm(0.881)=1.18 b) 1 – 0.92 = 0.08, therefore invNorm(0.08)=-1.41 Example 3: Example 4: Solution: Example 5: Solution: Example 6: Solutions: a) Therefore using the standard normal equations we get: and using systems of equations we can find that: b) The probability that the token will not work is p = 0.05. So this is a binomial probability: and we want to find , which is 0.7358.