hw3_sol

hw3_sol - 2.89 We first compute z—scores for each x...

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Unformatted text preview: 2.89 We first compute z—scores for each x value. _ xH—y 100—50 :2 a 'z— = 0' 25 . _ b. z= x—’u=1—_fi =—3 0' l c z: )c—;1:(J-—200=_2 a 100 d. ‘z= “again 0‘ 3 The above z—scores indicate that the )5 value in pan‘. a lies the greatest distancg above the mean and the x value of part b lies the greatest distance below the mean; 2.90 Since the element 40 has a 2—score of~2 and 90 has az—score of3 40-,u and = 90—# 0' a _2= :>—2(y=40—;z 330290—y 3y—20‘=40 3y+30190 :>p=40+2:r ' i By substitution, 40 + 20+ 30': 9O :>5cr= 50 i :> 0:10 By substitution, ,u = 40 + 2(10) : 60 l ' ‘ less ‘ ' . .00372 m /L, which IS 10“ was g at risk of ' . . ' 'vis cc the 90th percentile of the study sample In the subs; in the subdivision are not fl ,. o B a \ r4 '5‘ n m r. o H 3 _ r ‘ E (D C’ U‘ 11 n: 11> h— (D < (D ,_. o ,4, o .— K11 Q I. ‘g water with unhealthy lead levels. .296 a. From the problem, [u = 2.7 and a: .5 .7 x_ _ "I z=——# :zcr=x—,u:>x—,u+zcr 0' Forz:2.0,x=2.7+2.0(.5)=3.7 . . . . - Forz=—l.0, x=2.7 —1.0(.5)= 2.2 Forz = .5, x = 2.7 + 5(3) = 2.95 For 2 = —2.5, x = 2.7 — 2.5(.5) = 1.45 b. Forz = —l.6, x = 2.7 — l.6(.5) = 1.9 lfwe assume the distribution of GPAs is approximately mound—shaped, we can use the Empirical Rule. From the Empirical Rule, we know that @025 or z2.5% ofthe students will have GPAs in Pa P H g w 2“ above 3.7 (with z = 2). Thus, the GPA " p 1.7 2.2 2.7 3.2 3.7 corresponding to summa cum laude (top GPA ) will be greater than 3.7 (2 > 2). N ‘2 =\ We know that ~.16 or 16% ofthe students will have GPAs above 3.2 (z = 1). Thus, the limit on GPAs for cum laude (top 16%) will be greater than 3.2 (2 > I). We must assume the distribution is mound—shaped. 2,99 To determine if the measurements are outliers, compute the z—score. a ~x~;h65—57 _ 72 - Z — s H 11 — - 7 Since this z—score is less than 3 in magnitude, 65 is not an outlier. b _ x — E 2i — 57 _ , . _ z — S = 11 = -3.273 Since this z—score is more than 3 in magnitude, 21 is an outlier. _ x — E 72 — 57 . . c z — ~;—=—r = 1.364 Since this z—score is less than 3 in magnitude, 72 not an outlier. d Z = x — X = 9 — = 3_727 Since this z-score is more than 3 in magnitude, 93 s l] is an outlier. 2.101 The interquartile range is IQR = Q] — QL = 85 — 60 2 25. ‘ a ’ h t The lower inner fence = QL ~ 1.5(IQR) = 60 ~ 1.5(25) = 22.5. The upper inner fence = Q, + 1.5(1QR) = 83+ 1.5(25') =‘122.5. ’ The lower outer fence = QL — 3(IQR) : 60 —r 3(25) = —15. The upper outer fence = QU + 3(IQR) : 85 + 3(25) = 160. With only this information the box plot would look something like the followmg \ 1? \l + j—\ _ x + z + t l t + l l + 1 10 20 30 40 50 60 70 80 90 100 110 The whiskers extend to the inner fences unless no data pomts are that small or that large The upper inner fence is 122 5 However, the largest data pomt IS 100, so the whisker stops at 100 The lower inner fence is 22 5 The smallest data point is 18 so the whisker extends to 22 5. Since 18 IS between the Inner and outer fences it 13 designated with a *. We do not know if there IS any more than one data pomt below 22 5 so we cannot be sure that the box plot IS entirely correct. 2.102 a Median is approximately 4‘ b. QL is approximately 3 (Lower Quartile) Qu is approximately 6 (Upper Quartile) c. IQR:QU—QL~6—3=3 ' ' ' e is d The data set is skewed to the right since the right whisker IS longer than the left, ther one outlier, and there are two potential outliers. 500/ of the measurements are to the right of the median and 75% are to the left of the upper e. a quartile. f There are two potential outliers, 12 and 13. There is one outlier, 16. 3 1 a. Since the probabilities must sum to 1, P(E3)=1—P(El) P(E2) P(E4) P(E5):l .1~.2 .1 .1~.5 = _ « —PE —P(E)—P(Es) b. HE: ziwfiil .18 — ,1 :1 2P(E3) : .6 ::> P(E3) = .3 c. P(E3)=1—P(E1)—P(Ez)—P(E4)—P(E5):1~.1—.1~.1—.1:.6 \M ,p N 5 5' 5-43-21 120 35 a : :—\ ____ _ : :10 n 2 21(522)v 21-321 12 _ 3.7 C N_ 20‘ 20! % 20-19-13---3.2-1 ' n "‘5 ‘5!(20_5)r 5-4-32- :1. b. _ 2432902003 x 1018 =15 504 x , 1569209242 x 10” Ifwe denote the marbles as Bl, B2, R1, Each of the sample p ‘ omts would be equally likely. Thus, each woul of 1/10 ofoccum'ng. There is one sample point in A: (B1, There are 6 sample points in B: (B1, R1) Thus, P(B) : 6(ij I. 3. _§. 10 1 5‘ There are 3 sample points in C: (R1, R2) a 132) Thus, P(A) = i . 1-15‘14-13-w3-2~l R2, and R3, then the ten sample points are: (31,132) (Bl, R1) (B1, R2) (Bx, R3) (32 ,M@M@mmmmmww 10 J (Bl: R2) (B1, R3) (Bl, R1) (Bz, R2) (Bum) (R1, R3) (R2, R3). Thus, P(C) = 3[l—]6j= «(>1 ...
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This note was uploaded on 10/05/2008 for the course STAT 51 taught by Professor Singpurwalla during the Fall '08 term at GWU.

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hw3_sol - 2.89 We first compute z—scores for each x...

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