This article contains solutions to practice problems related to statistics. It covers topics such as sample size, random sampling, percentiles, variance, standard deviation, probability, and independence. The solutions are applicable to various subjects and courses offered by different colleges and universities.
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Question 1: Sample size = 100000/200 = 500 Question 2: k = (total sample)/(size of random sample) = 3500/175 = 20 sampling process will start between 1stand kthemployee (including both). Question3: Sorted List =[146, 204, 280, 298, 320, 356, 445, 450, 470, 786, 800, 820, 849, 918, 957, 964] 20thPercentile(index) = 0.20 * 16 = 3.2(index). Therefore 20thPercentile = 280. 60thPercentile (index) = 0.60 * 16 = 9.6(index). Therefore 60thPercentile = 786. Q1 is the median (the middle) of the lower half of the data, and Q3 is the median (the middle) of the upper half of the data Quartile Q1: 309 Quartile Q2: 460 Quartile Q3: 834.5 IQR = 834.5 – 309 =525.5 Question 4: a) Sample variance (s2)=(xi-x)2(n-1) xi= a number in data set x= mean (average) n = total number of samples in dataset In the given example, n = 8 x= (4+3+0+5+2+9+4+5)/8 = 4 Sample variance (s2) =(xi-x)2(n-1)=(0+1+16+1+4+25+0+1)/7 = 48/7 = 6.8571 b) Sample standard deviation (Sx) =(xi-x)2(n-1)=6.8571= 2.6186 Question 5: Using EmpericalRule,
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For 68% of data fall: Time Range 1=419 -27=392 Minutes. Time Range 2 = 419+27 = 446 Minutes For 95% of data fall: Time Range 1=419 –(2*27)=365 Minutes. Time Range 2 = 419+(2*27) = 473 Minutes For 99.7% of data fall: Time Range 1=419 –(3*27)=338 Minutes. Time Range 2 = 419+(3*27) =500 Minutes Question 6: n = 300 a) n(A) = 42+54+21 = 117 P(A) = n(A)/n = 117/300 = 0.39 b) n(Z) = 21+39+12 = 72 P(Z) = 72/300 = 0.24 c) P(A ∩ X) = 42/300 = 0.14 d) P(B ∩ Z) = 39/300 = 0.13 e) P(A∪C) = (n(A)+n(C))/n = (117+66)/300 = 0.61 f) P(A ∩ B) = n(A ∩ B)/n = 0/300 = 0
Question 7: DE A0.320.16 B0.200.12 C0.160.04 a.P(A) = 0.32+0.16 = 0.48 b.P(E) = 0.16+0.12+0.04 = 0.32 c.P(A∩E) = 0.16 d.P(A⎟E) =P(A∩E)/P(E) = 0.16/0.32 = 0.5 Question8: Let A = the event that a woman randomly selected participate in labor force, and B = the event that a woman randomly selected is married, and ~X = complement of X. a. P(A or B) = P(A) + P(B) - P(A and B) = .75 + .78 - .61 = .92 b. P[(A and ~B) or (B and ~A)] = P(A and ~B) + P(B and ~A) = P(A) - P(A and B) + P(B) - P(B and A) = .75 - .61 + .78 - .61 = .14 + .17 = .31 c. P(~A and ~B) = P(~(A or B)) = 1 - P(A or B) = 1 - .92 = .08 Question 9: a) P(B | A) =P(A∩B)/P(A) P(A∩B) = P(A)*P(B | A) = (0.40)*(0.25) = 0.1 b) P(A | C) =P(A∩C)/P(C) P(A∩C) = P(C)*P(A | C) = (0.35)*(0.80) = 0.28 c) If A and B are independent,P(A∩B) = 0 d)If A and C are independent,P(A∩C) = 0
Question 10: Ans: Variable 1 Variable 2EF A8575 B4055 C4085 D9525 a.P(F) = 240/500 = 0.48 b.P(B∪F) = 280/500 = 0.56 c.P(D∩F) = 25/500 = 0.05 d.P(B | F) =P(B∩F)/P(F) = (55/500)/(240/500) = 55/240 = 0.23 e.P(A∪B) = P(A)+P(B) = 255/500 = 0.51 f.P(B∩C) = n(B∩C)/n = 0 g.P(F | B) =P(B∩F)/P(B) = 55/95 = 0.58 h.P(A | B) =P(A∩B)/P(B) = 0 i.P(B) = 95/500 = 0.19 j. Based on your answers to these calculations, parts a, d, g and i, are variables 1 and 2 independent? Why or why not?I guess yes!