Download summary along with some simple problem solved

Things that you should know before starting these exercises
1. X - N(p,o) refers to a random variable X with a
mean of p and a standard deviation of o.
-
N(3.1 , 1.23!.refers to a random variable X
whose mean is 3.1and whose standard deviation is 1.23.1
[Thus, X
,..the normal with mean=0 and standard deviation=1".
[That is, Z - N(0 , 1) ]
STflNDAN
NqRrnflL
3.
e
Z-scores are in standard deviations from 0 (the mean).
[For example,Z=L.5 means 1.5 standard deviations from 0 (the mean].1
I^:o t= 1.5 iS l,S sfq..&*^t &q*'rqf i*rr {rm^ $\* il.er^ (0)
4.
The standard normal table ALWAYS gives the area
under the normal curve to the left of the given z value.
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7
The area under a normal curve represents a
probability-
*\^s G.r'rsu i s-.-0.X\ l]
,
(t^sn^g 9r* shdq"r
nsr#*\ 1"W*)
t=l
for example , PIZ < 1.2) is the probability that the standard
normal random variable Z is LESS THAN 1.2.
u,qq\l ({t'' ULlu)
t: l,I
So T@{ l.})= St\9
,:- ttsg 7.
/-a----
6. To use the standard normaltable to calculate
a probability
for a GREATER THAN, you must use the fact that
P(Z > a) =
1-
P{Z <a}.
for example, P(Z > 1] = 1- P(Z < 1)
...and you can look up P(Z < 1) in the standard normal table.l
[So,
(, **'F* t^t'Q*, P(e'i)= O't't
lJ
So 7(z,l)= t - o,tYl3 : 0,rstl
7.
To calculate the probability that a standard normal random
variable is between two numbers, you {a} look up the larger
number in the standard normal table, (b) look up the smaller
number and (c) subtract the smaller number from the larger.
P(t';?s6)
f6Th: zLr "rb)=?(z:1,'d")
,
l*rlcwp .tlr tc.}.-ks-,
o,ttcs
B.
-
o') I I*3 :O-t5
:
To calculate a probability related to a normal random
I
with mean p and standard deviation o
N(p, o] l, it must first be converted to a standard
variable
*
IX
ti
X
normal random variable Z using
, -f..
Once converted, thg probability is calculated
N ( 1$-?
l$ri
\n' N(
e-,-.pQa .. L'd.
Ld \n,'
Exa"-'pQ*'.
T 3.xz tl): f (ry
o.S
=f; G
9.
L? z- c.5 )
-
3
\
*\
(-t-4- L+(+\
=\ r(\:
LZ
r( e,,L."*(
(Z ':o'5)
o"6?t5
$c"'.-.\ttsl)
lf you must find a z-score corresponding to a certain probability,
I
z
1_
,/
s)
){ i5)* : Q"5\J
\).Jr,'3t)
$s 0\f^*.r;eul)*r) #'
\)o
./
-a-\
you have to find the number in the body of the standard normal
table that is closest to the given probability and then cross-
r ,1 t \
$t
Yka- T'>Tdi g1(tr
I f ,ni t\* Z-,1?ltJ*:\'|t-'+
f;a*6rqo
w\'r---Lsi"9**o'{1
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r fi$\*
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referencethatnumberbacktothecorrespondipgzvalue.
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Normal Distribution
Examples HoME
t-l
I NEXT I
Some simple problems to work in class (use table as needed):
1. Find the percent of area (round
to nearest percent) under the normal
curve
(a) between the mean and 0.35 standard deviations from the mean.
(b) between the mean and -0.35 standard deviations from the mean.
2. Find the percent of the total area under the standard normal curve
between the z-scores: z = -1.1 and z = 0.8.
3. Find the z-score such that 1 8% of the total area is to the (a) left and (b)
right of z.
4.P(-1.1 <Z<2.A1=7
5. lf X
- N(4.2,1.1),
(a) P(X < a)
(b) P(x > 5)
(clP(4<x<5)
calculate the following:
HOME
Some simple problems to work in class (use table as needed):
1. Find the percent of area {round to nearest percent) under the normal cun/e
N"k'. f*',s rnmsurej i"., :t*..tn,",[ dq{,rli.o.s
(a) between the mean and 0,Q€ standard deviations from the mean.
Su\-uq,Z=Olts
%;$*o,o\o'
A tb*A \*&.fs:
o,ir)- F(a^.il
bd'"**. b { b,\5 =?(acTt
| r*
luk..,1: 'in it6\,frui "*\'htlt
L
::1"o" Soco
- = e"bt6t
e=os;*-*d*oq*t
=
!
A"f".^, a:kt$.,ktfi""
A
f G.\,
l\;tr\$rt
=
t
lo:[i,.p
r
I
,
t
I
'
I
o"SocD
n
I fi
\l-otJ,
$
fof
r
n*,1
t
'/r tt
0rtr,'
t,
€=*l.l
t:o"E
lo"ky.;
{G
i^
si,i.
o;\Ki
= o.b5t1
Kr.it* G*t
13,6t %
n"f"l +o!l=
c.i
:
3
bKuqr.t
s1n p1fu;u
o"3C3a:
P(.1'.\)-P(t;i, l)
u,1
I
\otn t**{J,*
I
be1,t:t-66
*l,i{
q
lr
]qlr/q_
2. Find the percent of the total area under the standard normal curve
scores: z= -1.1 and z = 0.8.
.rl
I
nt"n*lc.urv(
P(q") - P(i.;o,ls)
h:1":11,. o,,
b*.*Y*,,..-4,3t
4O =
t:=(}-lt *:CI
7o
ir c ({5+ddn= t)
(b) between the mean and -0.35 standard deviations from the mean.
q6,.t,
(
A,39
o,
-r-'t, ?
39
35
Gr.cj kr-.*r,
Z.= -?:
--Ao
rt'='
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Normal Distribution Examples
t-NEXrl
Some simple problems to work in class (use table as needed):
(trr irn-r!\*p !>* k""1,u'J5,*y}-.,i
\*1 k*ror', fiq t4ittq
N.4u :. &*", \r$ia *d.v*,l- r,+ ru-$ +',* $; *,J T\1* Z se,;rt T\r* {1'vripntqs $\"Ls Q i t c,.
3. Find the z-score such that 18o/o of the total area is to the (a) lefl
f$**
(r)
(t;
E=i
sr,*o;nr;F[t"t:
..$-+trqi,l,
t^Li*"
a..,& $,,r,$ F"e S t\or**t,t(ou,
|
\ur,r,o\
fru.'- A- gr'*Q'
tn!"lq+ -]lj n.lt\t
.8,'.1\
*€.- :],1*
t,1
( l*rsjt**
)l',it
-+,
t*
r;-ff**"*"
ck:-*
4.P(-1.1 <Z<2.01=7
AG,. r;r)
f
f
'),SlSt rllbi1b
"
*.1{t\
"*'t
er:i\.
* iP{<, *l.i)
.,,*.r,t,*, \.u*
'**
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il,'t\ i)
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t,\I.r,
Normal Distribution Examples
Some simple problems to work in class (use table as
bI\
?=
vf
(a) P(X < 4)
f,(xzY)
=tr(*Aq*4) =fl(r.- *):fl(e.q,\t|b
T**^.\^Ut)Jt,r^"Y'.','-'
J-r*-.$J^n
*uil.^S\.!{ 9-"
rqqflVlnA0,\Jl?{ }:\ q) \
(b) P(x
'
.,i-1, f,,rta \"93ISU/cvlt"'^)
sl 1"!,&"5
"tE "
-+.)-\
(zaH3
?(x'i=TP(x.$ = \*?
T"l--l
--l P(e. *)= \- f (ea o")rl)- P(a. oJ 3) : - t"l tl, =@
I
^
r
\
\
^
J
\
I
I
\
{t*^\Lfu-
(c)P(4<X<5)
cz. 5*
n/
e0,l t P(24--o,tn)
:
L?.o^l)
t'G
o,\x
=f,f
t{ *"Hu- +
P(q.
xas)=P(t#&
\.\
l'1
@
-\J
n ;GI ]
61.".
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