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## M ARIO F . T RIOLA

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**STATISTICS**ELEMENTARY Chapter 4 Probability Distributions MARIO F. TRIOLA EIGHTH EDITION**Chapter 4Probability Distributions**4-1 Overview 4-2 Random Variables 4-3 Binomial Probability Distributions 4-4 Mean, Variance, Standard Deviation for the Binomial Distribution**4-1 Overview**This chapter will deal with the construction of probability distributions by combining the methods of Chapter 2 with the those of Chapter 3. Probability Distributions will describe what will probably happen instead of what actually did happen.**Combining Descriptive Statistics Methods and Probabilities**to Form a Theoretical Model of Behavior Figure 4-1**4-2**Random Variables**Definitions**• Random Variable a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure • Probability Distribution a graph, table, or formula that gives the probability for each value of the random variable**Table 4-1**Probability DistributionNumber of Girls Among Fourteen Newborn Babies x P(x) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0.000 0.001 0.006 0.022 0.061 0.122 0.183 0.209 0.183 0.122 0.061 0.022 0.006 0.001 0.000**Probability Histogram**Figure 4-3**Definitions**• Discrete random variable has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process. • Continuous random variable has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions.**Requirements for Probability Distribution**The sum of all the probabilities of the distribution equals 1. ΣP(x) = 1 where x assumes all possible values All probabilities of the distribution must fall between 0 and 1 inclusive. 0 P(x) 1 for every value of x**Mean, Variance and Standard Deviation of a Probability**Distribution Mean µ = [x•P(x)] Variance 2= [(x - µ)2 • P(x)] Standard Deviation = 2**Roundoff Rule for µ, 2, and**Round results by carrying one more decimal place than the number of decimal places used for the random variable x. If the values of xare integers, round µ, 2, and to one decimal place.**Example**A 4 point pop quiz was given and the scores ranged from 0 to 4, with the corresponding probabilities: 0.05, 0.2, 0.25, 0.3, 0.2 • Write the probability distribution in a table. • Verify whether a probability distribution is given. • Compute the mean, variance and standard deviation of the probability distribution.**Solution: Verify whether a probability distribution is given**• Yes it is a probability distribution because the sum of P(x) for all values of x is 1 and P(x) for all values of x is between 0 and 1, inclusive.**Solution: Compute the mean**• Enter x values into L1, Enter P(x) into L2. • Compute the products L3 = L1* L2. • Find the sum(L3) which is the Mean: μ = 2.4**Solution: Compute the variance and standard deviation.**• Having already found the mean of 2.4 • Compute L4 = (L1- 2.4)2* L2. • Find the sum(L4) which is the 2 = 1.34 1.3 • Find the which is the square root of 2 = = 1.16 1.2 • Mean: μ = 2.4 Variance: σ2= 1.3 Std Dev σ = 1.2**Definition**Expected Value The average(mean) value of outcomes, if the trials could continue indefinetly. E = [x • P(x)]**The Daily Number allows you to place a bet that the**three-digit number of your choice. It cost $1 to place a bet in order to win $500. What is the expected value of gain or loss?**E = [x • P(x)]**Event Win Lose x $499 - $1**E = [x • P(x)]**Event Win Lose x $499 - $1 P(x) 0.001 0.999**E = [x • P(x)]**Event Win Lose x $499 - $1 P(x) 0.001 0.999 x • P(x) 0.499 - 0.999**E = [x • P(x)]**Event Win Lose x $499 - $1 P(x) 0.001 0.999 x • P(x) 0.499 - 0.999 E = -$.50**Expected Value**• This means that in the long run, for each $1 bet, we can expect to lose an average of $.50. • In actuality a player either loses $1 or wins $499, there will never be a loss of $.50. • If an expected value is $0, that means that the game is fair, and favors no side of the bet.**Alternate Solution: Excluding the cost of playing.**Event Win Lose x $500 0 P(x) 0.001 0.999 x • P(x) 0.50 0 E = $.50, Which means the average win is $.50. If the game cost $.50 to play E = 0 and it would be a “fair” game and not favor either side of the bet. If the game cost $.75 to play E = -.25 and it would not be a “fair” game and would favor the house.