{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw3_sol

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ,[email protected]@mmmmmww 10 J (Bl: R2) (B1, R3) (Bl, R1) (Bz, R2) (Bum) (R1, R3) (R2, R3). Thus, P(C) = 3[l—]6j= «(>1 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online