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) = PX−µ
σ<89.5−89.21
0.17 =P(Z < 1.71) = .9564
(b) z= 0.52 ∵P(Z≤0.52) ≈.70
z=x−µ
σ; 0.52 = x−89.21
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(E2∪F2) = 221 + 118 + 138 + 654
1943 =.582
(c) P(F1) = 155 + 118
1943 =.141 ⇒(.141)2=.020
(d) P(E2∩F3) = 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(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 Ybe number of righthanded children. Then,
Yis binomial with n= 13 and π=.80.
⇒P(Y > 5) = 1 −P(Y≤5) = 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.