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Elementary Statistics - Final Exam 2 Solutions | MATH 220, Exams of Statistics

Material Type: Exam; Professor: Nashimoto; Class: ELEM STATISTICS [C3T1G1]; Subject: Mathematics; University: James Madison University; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 02/13/2009

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Math 220 : Elementary Statistics Kane Nashimoto
Exam 2 Solutions
Spring 2006
1. Normal random variable.
Note : X, octane level of gasoline, is normal with µ= 89.21 and σ= 0.17.
(a) P(X < 89.5) = PXµ
σ<89.589.21
0.17 =P(Z < 1.71) = .9564
(b) z= 0.52 P(Z0.52) .70
z=xµ
σ; 0.52 = x89.21
0.17 ;x= 89.21 + (0.52)(0.17) = 89.30
2. Correlation and regression by calculator.
(a) ˆy= 3.000 + 3.000 x
(b) r=.626
3. Probability.
E1:{selected student is female}F1:{selected student is self-supported}
E2:{selected student is male}F2:{selected student is parent-supported}
F3:{selected student receives financial aid}
(a) P(E1|F3) = 657
657 + 654 =.501
(b) P(E2F2) = 221 + 118 + 138 + 654
1943 =.582
(c) P(F1) = 155 + 118
1943 =.141 (.141)2=.020
(d) P(E2F3) = 654
1943 =.337 but
P(E2)P(F3) = 118 + 138 + 654
1943 ·657 + 645
1943 =.316
Two events are not independent.
4. Binomial random variable.
Note : X, number of lefthanded children, is binomial with n= 13 and π=.20.
(a) P(X < 3) = P(X2) = .5017
(b) P(6 X9) = P(X9) P(X5) = 1.000 .9700 = .0300
(c) “More than 5 righthanded” means “less than or equal to 7 lefthanded.”
P(X7) = .9988
Alternatively, let Ybe number of righthanded children. Then,
Yis binomial with n= 13 and π=.80.
P(Y > 5) = 1 P(Y5) = 1 .0012 = .9988
5. Correlation and regression by SPSS.
(a) ˆy= 308.354 + (0.124)(425) = 361.05
(b) r2=.008. Approximately 0.8% of the variability in the data is accounted for by
the linear relationship between Fall 2005 and Spring 2006 expenditures.

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Math 220 : Elementary Statistics Kane Nashimoto Exam 2 Solutions Spring 2006

  1. Normal random variable. Note : X, octane level of gasoline, is normal with μ = 89.21 and σ = 0.17.

(a) P (X < 89 .5) = P

X − μ σ

< 89.^5 −^89.^21

= P (Z < 1 .71) =. 9564

(b) z = 0. 52 ∵ P (Z ≤ 0 .52) ≈. 70 z = x^ −^ μ σ ; 0.52 = x^ −^89.^21

  1. 17 ; x = 89.21 + (0.52)(0.17) = 89. 30

  2. Correlation and regression by calculator. (a) ˆy = 3.000 + 3. 000 x (b) r =. 626

  3. Probability. E 1 : {selected student is female} F 1 : {selected student is self-supported} E 2 : {selected student is male} F 2 : {selected student is parent-supported} F 3 : {selected student receives financial aid} (a) P (E 1 |F 3 ) = 657 657 + 654

(b) P (E 2 ∪ F 2 ) = 221 + 118 + 138 + 654 1943

(c) P (F 1 ) = 155 + 118 1943

=. 141 ⇒ (.141)^2 =. 020

(d) P (E 2 ∩ F 3 ) = 654 1943 =. 337 but

P (E 2 )P (F 3 ) = 118 + 138 + 654 1943

⇒ Two events are not independent.

  1. Binomial random variable. Note : X, number of lefthanded children, is binomial with n = 13 and π = .20. (a) P (X < 3) = P (X ≤ 2) =. 5017 (b) P (6 ≤ X ≤ 9) = P (X ≤ 9) − P (X ≤ 5) = 1. 000 − .9700 =. 0300 (c) “More than 5 righthanded” means “less than or equal to 7 lefthanded.” ⇒ P (X ≤ 7) =. 9988 Alternatively, let Y be number of righthanded children. Then, Y is binomial with n = 13 and π = .80. ⇒ P (Y > 5) = 1 − P (Y ≤ 5) = 1 − .0012 =. 9988
  2. Correlation and regression by SPSS. (a) ˆy = 308.354 + (0.124)(425) = 361. 05 (b) r^2 = .008. Approximately 0.8% of the variability in the data is accounted for by the linear relationship between Fall 2005 and Spring 2006 expenditures.