Chapter 6 Review Solutions – Multiple Choice

1. Chapter 6 Review Solutions – Multiple Choice Solutions for Chapter 6 Review. If you find any errors, please e-mail me so I can correct them! lisam.weidman@cms.k12.nc.us Thanks!

2. Chapter 6 Review Solutions – Multiple Choice • B • C • B mean = np = (9)(1/3) • D std dev = • D P(X ≤ 1) = binomcdf (5,.2,1) • D P(X = 5) = binompdf (5,.2,5) • D P(X > 30) = 1 – binomcdf(100,.2,30) • C P(X ≤ 1) = binomcdf (9,1/3,1) • A P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – binomcdf (3,.8,1) • C 4 => (NYYY, YNYY, YYNY, YYYN) xP(success) xP(failure) • C

3. Chapter 6 Review Solutions – Free Response a. I would give you the table (probability distribution). I would NOT expect you to calculate these probabilities from the question. c. μx = ΣxiP(xi) = (0)(0.3038) + (1)(.4388) + (2)(.2135) + … = .9998 σx2 = Σ(xi– μx)2(P(xi)) = (0 - .9998)2(.3038) + (1 - .9998)2(,4388) + … σx2 = .7055 σx= √.7055 = .8399 **Be familiar with these formulas, but you may use the calculator to mind mean and standard deviation  Enter X values in L1 and P(X) values in L2. Use 1-Var Statistics L1, L2

4. Chapter 6 Review Solutions – Free Response (cont) d. P(X = 3) = 0.412 (from the table) e. P(X ≥ 3) = P(x = 3) + P(X = 4) = .0412 + .0026 = .0438 f. P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = .3038 + .4388 + .2145 + .0412 = .9973 1b – know how to draw a histogram (from Chapter 1)

5. Chapter 6 Review Solutions – Free Response 2.a. X = amount of money Fred wins b. *** 26/36  26 is the # of ways you lose (not 2, 3, 12,or 7) 6/36  6 is the # of ways you can roll a 7 4/36  4 is the # of ways you can roll 2, 3, or 12 c. E(X) = μx= 0(26/36) + 4(6/36) + 5(4/36) = $1.22

6. Chapter 6 Review Solutions – Free Response 3. a. Binomial  B: Binary Success – rolling 7, failure – not rolling 7 I: Independent Outcome of 1 roll does not effect other rolls N: Number of observations is fixed (8) S: Probability of success always the same (p = 6/36 = 1/6) b. X = # of 7’s that Don rolls c. Use your calculator binompdf (8,1/6,0) then change 0 to1, then 1 to 2…etc to populate probability distribution table: d. μx = np = 8(1/6) = 1.333…

7. Chapter 6 Review Solutions – Free Response • 3. e. P(X = 4) = binompdf(8,1/6,4) = .0260 • f. P(X ≥ 4) = 1-P(X < 4) = 1-binomcdf(8,1/6,3) = .0307 • g. P(X ≤ 4) = binomcdf(8, 1/6, 4) = .9954 • h. Chance of being within 1 standard deviation of the mean: • 1.3333 – 1.054 = .246 • 1.3333 + 1.054 = 2.354 • Binomcdf(8,1/6,2.354) – Binomcdf(8,1/6,.246) = .6326

8. Chapter 6 Review Solutions – Free Response • a. Geometric: • B: binary, two categories…success or failure • I: each shot is independent • T: she shoots until she makes it • S: p = .85 for each success • b. A trial is a free throw attempt. • X = the # of shots taken before she makes one. • c. P(X = 3) = (1-.85)2(.85) • or geompdf (.85,3) = .0191 • P(X ≤ 3) = geomcdf(.85,3) = .9966

9. Chapter 6 Review Solutions – Free Response • e. P (X>3) = (1 – p)n = (1-.85)3 • or 1 - P(X≤3) = 1 – geomcdf(.85,3) = .0034 • μx = 1/p = 1/.85 = 1.176 • g. • h. • * Not actually 0, but very close – rounds to 0 • ** Not actually one, but very close – rounds to 1

10. Chapter 6 Review Solutions – Free Response 5. M = Time it takes Sharon to get to work in the morning A = Time it takes Sharon to get home in the afternoon µM = 38 min σM = 7 min µA = 51 min σM = 5 min a. Mean of Sum of two Independent Random Variables = µT = µM + µA = 38 + 51 = 89 minutes b. Std Dev of the Sum of two Independent Random Variables = minutes

11. Chapter 6 Review Solutions – Free Response 5. M = Time it takes Sharon to get to work in the morning A = Time it takes Sharon to get home in the afternoon µM = 38 min σM = 7 min µA = 51 min σM = 5 min c. Mean of difference of two Independent Random Variables = µT = µA - µM = 51 - 38 = 13 minutes d. Std Dev of the diff of two Independent Random Variables = minutes