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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 ,.
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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 zscore 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' 54321 120
35 a : :—\ ____ _ : :10
n 2 21(522)v 21321 12 _ 3.7 C N_ 20‘ 20! % 2019133.21
' n "‘5 ‘5!(20_5)r 5432 :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 . 115‘1413w32~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 ...
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 Fall '08
 singpurwalla
 inner fence, upper inner fence, b. Forz, 2—score of~2

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