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Review 5.1 - 5.2 (regular Stats) - Random Variables Intro (Prob Distn) & Binomial Prob Ch. 5 (Random Variables): Review of 5.1 - 5.2 (1) a) Name: Find the missing value to make this a valid probability distribution x 0 1 2 3 4 p(x) 0.20 0.15 0.35 0.20 0.10 b) For the above probability distribution, calculate mx = E(X): E(X) = S xipi = 0.00 (2) 0.15 0.70 0.60 0.40 1.85 For the given values of the random variable X, and corresponding frequencies, complete the calculation of the standard deviation, sx Given: X 0 1 2 3 Freq 5 20 40 15 80 x p(x) 0 1 2 3 0.063 0.250 0.500 0.188 1.000 p(x)∙(x - m)2 0.205 0.165 0.018 0.264 x∙p(x) (x - E(x))2 0.00 0.25 1.00 0.56 3.29 0.66 0.04 1.41 mx = E(X) = 1.81 s2x = 0.652 = S pi∙(xi - m)2 S xipi sx = 0.808 (3) What are the 4 necessary conditions for using the Binomial Probability formula to calculate the probability of obtaining exactly k successes out of n trials • the number of trials, n, is fixed • each trial is independent • each trial is either a success (1) or a failure (0) • P(success) on each trial is the same (p) (4) The binomial probability of obtaining exactly k "successes" in n trials is given by: k n-k nCk ∙ p (1-p) a) what does p represent? p = probability of success on each trial b) what does (1-p) represent ? q = 1-p = probability of failure on each trial c) Write the binomial coefficient, nCk , using its equivalent factorial notation n! nC k = k!∙(n-k)! Example: 40C37 = 40! 37! ∙ 3! = 40 ∙ 39 ∙ 38 3∙2∙1 On TI-84 Calculator: 40 [math] [PRB] nCr 37 gives = C = 40 37 9880 9880 (5) For a given situation where the binomial probability distribution applies: The number of trials is 40 n = 40 The probability of a "success" is 3/8 and the probability of a non-success is 5/8. p = 0.375 Write a binomial probability expression involving the form, nCk ∙ pk(1-p)n-k, for: a) The probability of obtaining exactly 3 successes b) The probability of obtaining 7 or 8 successes C ∙ (.375)7(.625)33 + 40 7 c) The probability of obtaining 0 successes C ∙ (.375)3(.625)37 40 3 C ∙ (.375)8(.625)32 0 40 40C0 ∙ (.375) (.625) 40 8 d) The probability that you don't have exactly 20 successes 20 20 1 − 40C20 ∙ (.375) (.625) (6) Use your calculator to find each of the above probabilities in #5 On TI84 / TI84+ Calculator: [2nd] [Distr] BinPDF [enter] n , p , k a) BinPDF(40, 0.375, 3) = b) BinPDF(40, 0.375, 7) + 0.00357 c) d) BinPDF(40, 0.375, 0) = 0.0000146 BinPDF(40, 0.375, 8) = + 0.00884 = 0.01241 0.0000000068 1 − BinPDF(40, 0.375, 20) = << 0.000 000 006 842 1 − 0.0345 = 0.9655 Chapter Review (5.1 - 5.2) continued (7) Sam is a representative who sells large appliances such as refrigerators, stoves, etc. Let x = number of appliances Sam sells on a given day. Let f = frequency with which he sells x appliances. For a random sample of 240 days, Sam had the following sales record. x 0 1 2 3 4 5 6 7 f 9 72 63 41 28 14 8 5 240 Assume that the sales record is representative of the population of all sales days. a) Use the relative frequency to find P(x) for x = 0 to 7 x p(x) 1 0.300 2 0.263 3 0.171 4 0.117 5 0.058 6 0.033 7 0.021 5 6 7 Use a histogram to graph the probability distribution of part (a). 0.400 Relative Frequency = p(x) b) 0 0.038 0.300 0.200 0.100 0.000 0 1 2 3 4 x (# of appliances / day) 100% c) Compute the probability that x is between 2 and 5 inclusive (i.e., also including 2 and 5) Mutually Exclusive: P(2 or 3 or 4 or 5) = P(2) + P(3) + P(4) + P(5) Add: 0.263 0.171 0.117 0.058 = d) Compute the probability that x is less than 3 P(0) + P(1) + P(2) Add: e) 0.038 0.300 0.263 = 0.600 Compute the expected value of the x distribution. x p(x) f) 0.608 0 0.038 1 0.300 2 0.263 3 0.171 4 0.117 5 0.058 6 0.033 7 0.021 x ∙ p(x) = 0.000 0.300 0.525 0.513 0.467 0.292 0.200 0.146 Compute the standard deviation of the x distribution. s2x = S pi∙(xi - m)2 = Can use approach w/ #2 above p(x)∙(x - m)2 = 0.224 0.624 0.051 2.44 = S xipi 0.053 0.283 0.382 0.433 2.472 0.422 sx = sqrt( s2x ) = 1.572 Or, On Calculator: [Stat] [ Edit ] enter x-values in L1, enter p(x) values in L2 [Stat] [Calc] 1-Var Stats L1, L2 x-bar = 2.442 sx = 1.572 (8) The director of a health club conducted a survey and found that 23% of members used only the pool for workouts. Based on this information, what is the probability that, for a random sample of 10 members, 4 used only the pool for workouts? n = 10 ; p = 0.230 ; 10C4 ∙ (.375)4(.625)6 k=4 BinPDF(10, 0.230, 4) = 0.1225 (9) Of those mountain climbers who attempt Mt. McKinley, only 65% reach the summit. In a random sample of 16 mountain climbers who will attempt Mt. McKinley, what is the probability that: a) All 16 reach the summit 16C16 ∙ (.65)16(.35)0 Calculator: [2nd] [Distr] BinPDF [enter] n, p, k BinPDF(16, 0.650, 16) = b) At least 10 reach the summit 0 1 2 3 4 5 6 7 8 9 10 11 0.0010153 12 13 14 15 16 Recall, BinCDF (Cumulative Distribution Function) gives P( < k successes) always "less than or equal to" CDF for < On Calculator: [2nd] [Distr] BinCDF [enter] n, p, k P(at least 10) = P( k > 10) = 1 − P( k < 9) = 1 − BinCDF(16, 0.650, 9) = 1 − 0.3119 = 0.6881 c) No more than 12 reach the summit 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 13 14 15 16 P(no more than 12) = P( k < 12) = = BinCDF(16, 0.650, 12) = 0.8661 d) From 9 to 12 reach the summit (including 9 & 12) 0 1 2 3 4 5 6 7 8 9 10 11 12 subtract CDF(… 8) Can add Individual probabilities =BinPDF(16, .65, 9) + BinPDF(16, .65, 10) + BinPDF(16, .65, 11) + BinPDF(16, .65, 12) 0.1524 + 0.1982 + 0.2008 Or can use Cumulative probabilities P( 9 < k < 12) = P( k < 12) − P (k < 8) = BinCDF(16, 0.650, 12) − = 0.8661 − + 0.1553 = 0.7067 BinCDF(16, 0.650, 8) 0.1594 = 0.7067