DISTRIBUTIONS RANDOM VARIABLESExample - You have a bag with 8 oranges and 3 of them are rotten. Choosea sample of 3. Let X be the number of rotten oranges in the sample. Writea probability distribution table for the number of rotten oranges in the sample.Graph the probabiltiy distribution histogram.We will need to use counting to find the probabilities. The probability of event EisP (E)=n(E)n(S)=number of outcomes in Enumber of outcomes in SFind the probability distribution table:eventX P(X)0R,3G 0 C(3,0)C(5,3)/C(8,3) = 10/561R,2G1 C(3,1)C(5,2)/C(8,3) = 30/562R,1G2 C(3,2)C(5,1)/C(8,3) = 15/563R,0G3 C(3,3)C(5,0)/C(8,3) = 1/56Put this into a histogram.We will have X on the horizontal axis and the probability P (X) on the verticalaxis.We have a rectangle 1 unit wide centered on each X value and its height is theprobability:2Example - We are given the following data for the number of a certain magazinesold per week for the past year at a newstand:number soldfrequency probability15 516 417 818 1019 820 1521 2total 52Find the probabililty distribution histogram for this data set:3EXPECTED VALUEHow many rotten oranges would you expect to get in your sample of 3? Howmany magazines do you expect to sell? We have a formula for the expected valueof a probability distribution:E(X)=X1P(X1)+X2P(X2)+... + XnP (Xn)Look at the oranges:E(X)=0·1056+1·3056+2·1556+3·156=6356=1.125We can show this on the histogram - this is where it would “balance”.4Look at the magazines,E(X)=15·552+16·452+17·852+18·1052+19 ·852+20·1552+21·252=18.25What is the “average” number of magazines sold?One kind of average is the MEAN,¯X,¯X =number of magazinesnumber of weeks=15 · 5+16·4+... +21·252=18.25Same as the expected value! So we can let the calculator do the work.5There are two other kinds of “averages”.The MEDIAN is the number in the middle if all the data is lined up in order.The calculator will list this for you when you do the 1-Var Stats.The other is the MODE.This is the data that occurs most often. You must see the frequency distribution tofind this.A set of data may have no mode, one mode or more than one mode.The mode will work for non-numerical data too.6ODDSIf the probability that event E occurs is P (E), we say that the ODDS IN FAVOROF E areP (E)1 − P (E)=P (E)P (Ec),P(E)6=1The ratio is usually expressed as integers.example - what are the odds in favor of rolling doubles with two fair dice?E is the event that we roll doubles,so P (E)=6/36Ecis the event that we don’t roll doubles,so P (E)=30/36The odds are thenP (E)P (Ec)=6/3630/36=630=15So we would say the odds in favor of rolling a double are 1to5 or 1:5.If we are given the odds, we often want to change this to probability.If the odds in favor of an event are a : b, then the probability of the event isP =aa + bexample - we are given the odds in favor of a certain horse winning a race are6to5. What is the probability that the horse will win?P =66+5=611≈ .557Variance and Standard DeviationThe variance and the standard deviation measure the spread in your data.We can find these values for raw data, grouped data and probability distributions.The work will be done by the calculator.Example - You have a class of 10 students. The students take a 20 point quiz andthe scores are:20, 13, 15, 19, 18, 17, 14, 16, 16, 18Find the mean, median, mode, standard deviation and variance of this data set.Answer - Put the data values into a list on the calculatorWe then findmean = µ = 16.6 (it is a population)median = 16.5standard deviation = σ = 2.1071To find the mode, look what score appears the most often.16 and 18 each appear twice, so this data is bimodal,modes = 16 and 18To find the variance, σ2, you can use the formula or square the standard deviation,variance = σ2= 2.10712= 4.44008Example - Find the mean, median, mode, standard deviation and variance for thenumber of seeds in a grapefruit from a sample of 50 grapefruits:number of seedsfrequency3 548515612710the mean is ¯x =5.28 (it is a SAMPLE)the median is 5the mode is 5the standard deviation is S =1.24619Example - find the mean and standard deviation for the number of hearts in ahand of 6 cards. Draw the histogram for this distribution and show the mean andstandard deviation on the histogram.Start with a table for the probability.We will have P(E)=n(E)/n(S):number ofprobabilityhearts0 C(13, 0)C(39, 6)/C(52, 6) ≈ .16021C(13, 1)C(39, 5)/C(52, 6) ≈ .36772C(13, 2)C(39, 4)/C(52, 6) ≈ .31513C(13, 3)C(39, 3)/C(52, 6) ≈ .12844C(13, 4)C(39, 2)/C(52, 6) ≈ .02605C(13, 5)C(39, 1)/C(52, 6) ≈ .00256C(13, 6)C(39, 0)/C(52, 6) ≈ .0001the mean = µ =1.5002 (a probability distributions is ALWAYS a population)the standard deviation = σ =1.007Draw the histogram:10Chebychev’s TheoremWe can ESTIMATE the probability that data lies within k · σ of the mean of apopulation with Chebychev’s theorem:P (µ − kσ ≤ X ≤ µ + kσ) ≥ 1 −1k2Example - You have a probability distribution with a mean of 100 and a standarddeviation of 5.a) Estimate the probability that an outcome of the experiment will be within 5standard deviations of the mean.b) Estimate the probability of an outcome of an experiment is between 80 and120.c) Find a value c such that guarantees that the probability is at least 75% that anoutcome of the experiment lies between 100 − c to 100 +
View Full Document
Unlocking...