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 DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full 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 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 ,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 ...
View
Full
Document
This note was uploaded on 10/05/2008 for the course STAT 51 taught by Professor Singpurwalla during the Fall '08 term at GWU.
 Fall '08
 singpurwalla

Click to edit the document details