STATISTICS TEST # 1 SOLUTIONS |
SOLUTIONS ARE IN [ square brackets ] (mark values in parentheses)
1/
2/ Give the number of ways a student can mark her answers
3/ 6 baseball teams. how many ways can:
4/ 2 cards are drawn from a well-shuffled deck of 52 playing cards. (6)
5/ 15 tapes she wants to select 6. (4)
6/ 4 accountants, 2 financial planners from 8 accountants, 5 financial planners: (2)
b) How many 6-person teams can be hired if no attention is paid to profession? [ 13 C 6 = 1716 ]
(1)
7/Explain in words what events are represented by the regions numbered: a) Harry graduates but doesn't get a car. (7)
8/ P(R) = 0.62, P(T) = 0.75, and P(R 3 T) = 0.43, find:
(4)
9/ If one of these terms is selected at random, find each of the following probabilities:
(6)
10/ P(R)= 0.4, P(C) = 0.5, P(R 3 C) = 0.25.
a) P(C 3 R) / P(R) = P(C / R) = 0.25/0.40 = [ 0.625 ] (8)
11/ A set of data yield the following: n = 12, Sx = 192, and Sx2 = 3,248. µ the mean = Sx/n = 16 r the standard deviation = (3)
TOTAL (50)
(3)
to a multiple choice test if there are:
a) 8 questions 3 choices [ 3 8 = 6561 ](1)
b) 3 questions 8 choices [ 8 3 = 512 ] (1)
c) 10 questions 4 choices for 1st - 3; 5 choices for last - 7 [ 4 3 $ 57 = 5,000,000 ] (2)
a) 3 teams finish in 1st, 2nd and 3rd place? [ 6 $ 5
b) all 6 teams finish 1st through 6th? [ 6! = 720 ] (2)
What are the probabilities of getting:
a) 2 jacks?
b) 2 clubs?
c) an ace and a king?
In how many ways if:
a) the order matters? [ 15 P 6 = 3, 603, 600 ]
b) the order doesn't matter? [ 15 C 6 = 5, 005 ]
a) How many 6-person teams can be hired under the given conditions? [ 8 C 4 $ 5 C 2 = 700 ]
a) 1
b) 2
c) 3
d) 4
e) 3 & 4
f) 2 & 3
g) 1 & 2 & 3
b) Harry graduates and gets a car.
c) Harry gets a car but doesn't graduate.
d) Harry doesn't graduate and he doesn't get a car.
e) Harry doesn't graduate this year.
f) Harry gets a car this year.
g) Harry graduates and/or gets a car.
a) P(R') = [ 1 - P(R) = 0.38 ]
b) P(R 4 T) = [P(R) + P(T) - P(R 3 T) = 0.94 ]
c) P(R 3 T ') = [P(R) - P(R 3 T) = 0.19 ]
d) P(R 4 T ') = [P(R) + P(T ') - P(R 3 T ') = 0.68 ]
a) P(M) = 625/1000 = 0.625
b) P(I') = 300/1000 = 0.3
c) P(M'/Z) = 175/300 = 0.583
d) P(F/I) = 200/700 = 0.286
e) P(Z/M') = 175/375 = 0.467
f) P(F'/Z') = 500/700 = 0.714
b) P(R / C) = 0.25/0.50 = [ 0.5 ]
c) P(R 4 C) = 0.4 + 0.5 - 0.25 = [ 0.65 ]
d) If P(R) $ P(C) = P(R 3 C), the events are independent.
But 0.20 ! 0.25 so [ R and C are not independent events. ]
Find:
r2 the variance =
.
Back to Statistics MathRoom Index
.