Casio CFX-9800G - Page 73

To calculate the deviation of
the unbiased variance, the
difference between each da-
tum,-and-mean of
data.-
the above e
Determine the- following:
P distribution
Q distribution ~'
R distribution
t distribution
To calculate x and on-1 for
the following data
Class no.
To determine Med, Max and «.;
Min.
Frequency.
(Continuing) (6i1)(Fa] (DEV)
F3)(xon 1) AES
55 GIF) (x) Bs):
Sale) es |
518 (x=)
» FN(P()0.20) 68
F2)(Q() 0.25068
F3}(R() 30) 69
Fat) 580)
(Fs)(SET) FJ (STO) BM
Ga) GF) (Scl) BES) (CAL)
: ~ EO) (Med)
© {Clears memory)
1103)(;) 1069 (07)
1306)(;)310(DT)
15063)(;) 24) (DT)
170(F)(DT)FA(DT)
_ 190E)(DT)EN(DT)ER(DT)
F6)(CAL) G5 (£) 3 (n)
- (IFA (DEV) Fi (x es
F3)(xon—-1)&2)
[Fs] (Max) Be
[F4) (Min) 2
@miEa(PaR)
1.982142857
. 0.57926
0.098706
1.352-03
3.51188458428
110
130
150
170
190
: 7@
137.7142857
18.42898069
138
190
110
Example Operation _ Display’.
The table below shows the Fs}(SET) (STO) a)
heights
of
20
college
stu-
F2)(EDIT)F3)(ERS)
(YES)
ve
dents. Determine what per- :
centage of the students fall ms 158.5(](D7) 158.5
in
the
range
160.5
cm
to
le
160.59
(DT)
:
160.5
175.5
cm.
Also;
in
what
per-
“|
*
a
S
centile
do
the
175.5
cm
tall
163.365(;)2E(DT)
os
163,3
students
fall?
167.5@3(;)2Q(DT)
|:
s
|
167.5
a Soon = 170.23)(;)3G9(DT) 170.2
= a. one a 173.303(;)4@(D7) 173.3
2 160.5 1 175.5(3)(;) 2 (DT) 175.5
3 163.3 2 178.6 §3)(;) 29 (DT) 178.6
4 167.5 2 180.4 63)(;)2(DT) as 180.4
5
170.2
3
ra
186.7F(DT)
|
oie
186.7
6 173.3 4 ae 2 (POR) Ls Se ie
7
175.5
ae:
(Normalized
variate
tfor
160.5
cm)
|
8 178.6 2 Ht() 160.5069); —1.633855948
9 180.4 2 : i (5 — 1.634).
veal 175.5006 | 0.4963343361.
10 186.7 1 9D. : \
; : (=0.496)
(Percentage of total) ay : :
©(P()0.4960)3 |" :
€0(P( )@)1.6340 jeg 0.638921,
ts (Percentile) | (Result: 63.9% overall)
&3)(R( )0.4960) 68 0.30995
‘ (Result: 31 percentile)
*The
following
cietnbubony
curves
illustrate
the
two
concepts
coveréd
i
in
this
problem,
eFercentage of the Total
_ 71.634 0.495 y csatke y+ 10096 1684"
Percentile