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Midterm Exam Solutions for Statistical Methods | STAT 51100, Exams of Data Analysis & Statistical Methods

Purdue University Data Analysis & Statistical Methods

Prof. Mihails Levins

Material Type: Exam; Professor: Levine; Class: Statistical Methods; Subject: STAT-Statistics; University: Purdue University - Main Campus; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 07/30/2009

koofers-user-0xp 🇺🇸

10 documents

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111

()

nnn

iii

ax b nb a x a x

baxb

nnn

===

===+=

∑∑∑

211

22 2

() ( ( )

[( )] ( )

)

yaxbaxb

snn

ax x a x x

−+−

−−

===

−−

∑∑

50 551 452

012

543

( 3) 1-[p(x=0)+p(x=1)+p(x=2)]

= 1-[b(0;5,0.9)+b(1;5,0.9)+b(2;5,0.9)]

= 1-[( ) 0.9 (1 0.9) ( ) 0.9 (1 0.9) ( ) 0.9 (1 0.9) ]

1 [1 10 4.5 10 8.1 10 ] 0.9914

−−−

≥=

××− +××− +××−

=− × + × + × =

Let W be withdrawal petition, C be course substitution request

646

642

10 10

() ()()

(6 ) (4 2 ) () ()

11516

7.62%

210 210 210

p p W p Cand W=+ =+

=++=

6453 4635

()()

10 9 8 7 10 9 8 7

0.0714 0.0714 14.28%

p p WCWC p CWCW ⎛ ⎞⎛⎞⎛⎞⎛⎞⎛ ⎞⎛⎞⎛⎞⎛⎞

=+ = +

⎜ ⎟⎜⎟⎜⎟⎜⎟⎜ ⎟⎜⎟⎜⎟⎜⎟

⎝ ⎠⎝⎠⎝⎠⎝⎠⎝ ⎠⎝⎠⎝⎠⎝⎠

=+=

Discover Exams of Data Analysis & Statistical Methods Purdue University

Partial preview of the text

Download Midterm Exam Solutions for Statistical Methods | STAT 51100 and more Exams Data Analysis & Statistical Methods in PDF only on Docsity!

1 1 1

n n n i i i i i i

ax b nb a x a x y b ax b n n n

= = =

2 2 2 1 1

2 2 2 1 1 2 2

[ ( )] ( )

n n i i i i y n n i i i i x

y y ax b ax b ) s n n a x x a x x a s n n

= =

5 0 5 5 1 4 5 2 0 1 2 5 4 3

( 3) 1-[p(x=0)+p(x=1)+p(x=2)] = 1-[b(0;5,0.9)+b(1;5,0.9)+b(2;5,0.9)] = 1-[( ) 0.9 (1 0.9) ( ) 0.9 (1 0.9) ( ) 0.9 (1 0.9) ] 1 [1 10 4.5 10 8.1 10 ] 0.

p x

− − −

× × − + × × − + × × −

= − × + × + × =

Let W be withdrawal petition, C be course substitution request

6 4 6 6 4 2 10 10 6 6

(6 ) (4 2 ) ( )^ ( )( )

p = p W + p Cand W = +

b. 6 4 5 3 4 6 3 5 ( ) ( ) 10 9 8 7 10 9 8 7 0.0714 0.0714 14.28%

p = p WCWC + p CWCW = ⎛⎜^ ⎞⎛⎟⎜^ ⎞⎛⎟⎜^ ⎞⎛⎟⎜^ ⎞⎟^ +⎛⎜^ ⎞⎛⎟⎜^ ⎞⎛⎟⎜^ ⎞⎛⎟⎜^ ⎞⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ = + =

The average amount of damage the company has to cover after the deductible was paid is 3 1

( (^) i 500) ( (^) i ) 0*0.8 (1000 500) 0.1 (5000 500) 0.08 (10000 500) 0.02 600 i

x px

∑ −^ ⋅^ +^ =^ −^ ⋅^ +^ −^ ⋅^ +^ −^ ⋅^ =

Premium = Profit + The average amount of damage = $200 + $600 = $