1. For each problem also draw a picture of the normal curve and shade the area you have to compute
Let Z represent a variable followmg a standard normal distribution.
(a) F ~ n dthe proportlon that
1s less
than z=2 00
z-
k&
1s
-
between
aka
13 and
Find the proportlon that 1s greater than z=l 86
b&n7%f
af 1.86 CL-
+o -C&
b@$ -
= 1 75
-
o,si\b =
&
k=2.asr
F O h ~ Y iO
&t 7 1 1 L .
t=z
(b) Find the Proportlon that
(c)
cmWs-
5e(0~
JiPs
h a t-sate
t=-.13
6 -
b~,
8 b ) =(-.q&&,
$1
t-tb7S
I , ~ ' O : 0.03\4
id) Flnd the z-score for the 64th percentile W e w w l-w
9~ atea
af
.64
7~ n c d
&=r
Bh
ole
cotreyor6L -to
.La
- -teg
ThSiBP
(e) Flnd the z-scores that bound the rn~ddle50% of all data
61
i f ) F ~ n dthe z-score for the 24th percent&
%a 9a.u
crs
?-Scot+?
d)
I L ~
=
w
c
4
LX
-
all
What
QS~W
is
ah.
LW
ar-eae
- 811I
2 Former ISU basketball layer Kelv~nCato
(a)
%=?
IS83
~nchestall
tus correspond~ngz-score7
a u-
Fz7.3
b=3 =
)
t-sL0r-e =
- 4.33
83-70
3
=
(b) What praport~onat men are taller than him?
4,33 is
(
O# 8 . q
&-f.~d
b~
( I
~Lyi
LL
/&
1 ;S
~YI
3.+4 i A
- P I ~ ~ U O ~ )
a I h d
~~ ,30484
3 - 0 - 0 ~ ~4 33 is
+ brhe(ey k b o d y i a l ( ~ rJ0-k
a
-
RW
kim .
xu(a
4.33
3. Since the length of a downhill ski is related to the height of tho individuals renting them, it is Fair
t o assume that a normal distribution would describe the length of women's skis at rental outlets in
Colorado. The mean of the distribution is 150 cm and the standard dev~ationis 10 cm.
f
(a) What is the proportion of women's ski lengths that are less than 130 cm?
,--i--.
2 = 133 - 1 5 ~
1
0
I
0,0228
(b) What
2,
IS
a-
R s -1!3
=
(c) What
is
130 \&
the proportion of women's s k ~lengths that are greatcr t b n 125 cm?
&
-2,s
-
I - , 0062.= 0 , 9 7 3
the proport~onof women's s k ~lengths that are between 125 and 1557
.
(0
= G 9 1 5 - 0,0062=0 , Q G J
+
05/.-i.-\%
women's slu a rental shop should carry so that only
percent of the c o s t ~ m e r sw'll ask to rent
.2 .
a longcr ski?
had
to do b = d - w a r d s d ~ k h ' <
~ ?, .,yd
,,,
I - i S c = e (k-I&42%
+
% ' ~ b + ~ =
v - k b ; L = 2.
-. - - -...-.
X=>2<05 .(a+ls0=(7Q,S
...... .. . .
... ... ..-- . ~
-
-
.
4. The BMI for m a l a age 20 t o 74 is follows approximately a normal distribution with mean p = 27.9
and standard deviation o = 7.8. Use the 68-9699.7 rule to find
.
E*
(aj the percentage of males with BMI less than 20.1.
hckki
L;4
20~1is I
F bdoa
kid&
68q3
d
32-93 9-
Uak)
b,d fo
w i
169,
(bj the percen a g s of males w ~ t hBMI gr ater -than
- . 3 . 21
j ' ~
12.3
=>
3
~ ~ e , p o ~4-ad =2 6 be lo^
9 7 . 9 3 0 -be
a
'
-
-16
+I k-
.(ep ~ c e a k
23,9 + 3*?.8= 51,
(d) the value such that 0.15% of rnalcs have BMI's greater than the valne.
(-
51.3
44173
-
-3 03c
h n , i e i& L;&&
51.3