Math 220 : Elementary Statistics Kane Nashimoto
Exam 2 Solutions
Fall 2006
1. Probability.
E:{selected student is male}
F:{selected student prefers morning classes}
Note : P(E) = .46 ; P(F) = .67 ; P(E∩F) = .29
(a) P(E∪F) = P(E) + P(F)−P(E∩F) = .46 + .67 −.29 = .840
(b) P(F|E) = P(E∩F)
P(E)=.29
.46 =.630
(c) P(at least one male) = 1 −P(all female) = 1 −(.54)3=.843
(d) P(E∩F) = .29 but P(E)P(F) = (.46)(.67) = .308
⇒Two events are not independent.
2. Normal random variable.
Note : X, heating cost, is normal with µ= 450 and σ= 15.
(a) P(X > 425) = PX−µ
σ>425 −450
15 =P(Z > −1.67)
= 1 −P(Z≤ −1.67) = 1 −.0475 = .9525
(b) z= 1.28 ∵P(Z≤1.28) ≈.90
z=x−µ
σ; 1.28 = x−450
15 ;x= 450 + (1.28)(15) = 469.20
3. Correlation and regression by calculator.
(a) r=.245
(b) ˆy= 61.934 + 0.286 x
4. Binomial random variable.
Note : X, # of customers ordering Daily Special, is binomial with n= 20 and π=.25.
(a) P(X < 5) = P(X≤4) = .4148
(b) P(X= 7) = P(X≤7) −P(X≤6) = .8982 −.7858 = .1124
(c) P(2 < X < 10) = P(X≤9) −P(X≤2) = .9861 −.0913 = .8948
(d) “More than 15 not order Daily Special” means “4 or fewer order Daily Special.”
⇒P(X≤4) = .4148
Alternatively, let Ybe # of customers not ordering Daily Special. Then,
Yis binomial with n= 20 and π=.75.
⇒P(Y > 15) = 1 −P(Y≤15) = 1 −.5852 = .4148
5. Correlation and regression by SPSS.
(a) r2=.190
(b) se= 2.618
(c) ˆy= 28.278 + (1.667)(1.42) = 30.645