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 1.B 2.C 3.B mean = np = (9)(1/3) 4.D std dev = 5. D P(X ≤ 1) = binomcdf (5,.2,1) 6.D P(X = 5) = binompdf (5,.2,5) 7.D P(X > 30) = 1 – binomcdf(100,.2,30) 8.C P(X ≤ 1) = binomcdf (9,1/3,1) 9.A P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – binomcdf (3,.8,1) 10.C 4 => (NYYY, YNYY, YYNY, YYYN) xP(success) xP(failure) 11.C

3. Chapter 6 Review Solutions – Free Response 1.a. I would give you the table (probability distribution). I would NOT expect you to calculate these probabilities from the question. c. μ x = Σx i  P(x i ) = (0)(0.3038) + (1)(.4388) + (2)(.2135) + … =.9998 σ x 2 = Σ(x i – μ x ) 2 (P(x i )) = (0 -.9998)2(.3038) + (1 -.9998)2(,4388) + … σ x 2 =.7055 σ x = √.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 X01234 P(X).3038.4388.2135.0412.0026

4. Chapter 6 Review Solutions – Free Response 1.(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) = μ x = 0(26/36) + 4(6/36) + 5(4/36) = $1.22 X$0$4$5 P(X) ***26/366/364/36

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. μ x = np = 8(1/6) = 1.333… X012345678 P(X).2326.3721.2605.1042.0260.0042.0004.0000

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 4.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 d.P(X ≤ 3) = geomcdf(.85,3) =.9966

9. Chapter 6 Review Solutions – Free Response 4.e. P (X>3) = (1 – p) n = (1-.85) 3 or 1 - P(X≤3) = 1 – geomcdf(.85,3) =.0034 f.μ x = 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 X12345678910 pdfP(X).85.1275.0191.0029.00040* cdfP(X).85.9975.9966.9995.99991**

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 µ M = 38 min σ M = 7 min µ A = 51 min σ M = 5 min a. Mean of Sum of two Independent Random Variables = µ T = µ M + µ A = 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 µ M = 38 min σ M = 7 min µ A = 51 min σ M = 5 min c. Mean of difference of two Independent Random Variables = µ T = µ A - µ M = 51 - 38 = 13 minutes d. Std Dev of the diff of two Independent Random Variables = minutes