STATISTICS TEST # 1 SOLUTIONS

SOLUTIONS ARE IN [ square brackets ] (mark values in parentheses)

1/

(3)

2/ Give the number of ways a student can mark her answers
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)

3/ 6 baseball teams. how many ways can:
a) 3 teams finish in 1st, 2nd and 3rd place?
[ 6 \$ 5
\$ 4 = 120 ]
b) all 6 teams finish 1st through 6th?
[ 6! = 720 ] (2)

4/ 2 cards are drawn from a well-shuffled deck of 52 playing cards.
What are the probabilities of getting:
 a) 2 jacks? b) 2 clubs? c) an ace and a king?(6)

5/ 15 tapes she wants to select 6.
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 ]

(4)

6/ 4 accountants, 2 financial planners from 8 accountants, 5 financial planners:
a) How many 6-person teams can be hired under the given conditions?
[ 8 C 4 \$ 5 C 2 = 700 ]

(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) 1 b) 2 c) 3 d) 4 e) 3 & 4 f) 2 & 3 g) 1 & 2 & 3

a) Harry graduates but doesn't get a car.
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.

(7)

8/ P(R) = 0.62, P(T) = 0.75, and P(R 3 T) = 0.43, find:
 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 ]

(4)

9/ If one of these terms is selected at random, find each of the following probabilities:
 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

(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 ]
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. ]

(8)

11/ A set of data yield the following: n = 12, Sx = 192, and Sx2 = 3,248.
Find:

µ the mean = Sx/n = 16
r2 the variance =

r the standard deviation =

(3)

TOTAL (50)

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