a)
7! / 3! = 840Q1.
a)
|
Stem |
Leaf |
Frequency |
|
1 2 3 4 5 6 7 8 10 |
9 6 8 9 0 0 2 4 5 5 6 7 7 7 9 9 9 1 1 1 2 2 3 3 5 5 7 8 9 1 3 6 7 1 4 4 0 3 5 1 |
1 3 13 12 1 3 3 3 1 |
|
Total: |
40 |
|
Note: 1|9 means 19.
b)
Mean=47.6
Variance=353.43
c)
First quartile=(35+36)/2=35.5
Second quartile=(41+42)/2=41.5
Third quartile=(51+63)/2=57
d)
Range =101-19=82
Inter-quartile range=57-35.5=21.5
e)
f)
Positively skewed distribution
Q3.
a) 10P6 = 151200
b) 106 = 1000000
Q4
a) 6! = 720
b) 3! = 6
c) 3! 2! 3= 48
d) 3! 3! 2! = 72
Q5
a)
7! / 3! = 840
b) 8! / 2! 3! = 3360
c) 9! / 2! 2! = 90720
d) 11! / 4! 4! 2! = 34650
Q6
Cast1, there are three ‘b’,
3P0 =1
Cast 2, there are two ‘b’,
3P1×3 =9
Cast 3, there is one ‘b’,
3P2×3 =18
Cast 4, there is no ‘b’,
3P3 = 6
There are (1+9+18+6)= 34 number of permutations.
Q7
a) 8P2×4P1 = 224
b) 8C2×4C1 = 112
Q8
10C3×4C2 = 720
Q9
The total number of beads = (21+9+11+9)= 50
a)
21 / 50
b) 9 / 50 +9 /50 = 9 / 25
c) 1- (21 / 50 + 9 / 50) = 2 / 5
Q10
a) 9 / 12 × 8 /11 = 6 /11
b) 2C1 (3 / 12 × 9 /11) = 9 /22
c) 3 /12 × 2 /11= 1 /22
Q11
The total number of cakes (6+4)=10.
The probability that the waiter will make a mistake
=1-the probability he will get the right cake (both chocolate cake)
=1-(6/10)(5/9)
=2/3
Q12
The probability that a bridge will not close = 1-0.2=0.8
a) The probability that we can go from Town A to C
=1-the probability that we cannot go from Town A to C
=1-(the probability that the both Bridges X and Y close or
the probability that the both Bridges W and Z close but Bridges X and Y open or the probability that the both Bridges W and Z close but Bridges X or Y open)
=1-[(0.2×0.2)+ (0.2×0.2) (0.8×0.8)+ (0.2×0.2) (0.8×0.2
=1-0.0784
=0.9216
b) The probability that we can go from Town A to C
=1-the New Bridge close and
the probability that we cannot go from Town A to C with no New Bridge
=1-0.2(0.0784)
=0.98432
Q13
a) 4! ×6! =17280
b) The total ways of sitting – the ways that four girls sit together
=9! -4! ×6!
=345600
Q14
a) (1/6) (1/6) (1/6)=1/216
b) 3C1(3/6) (3/6) (3/6)=3/8
c) 1-(5/6) (5/6) (5/6)=91/216
d) 3C1(3/6) (3/6) (3/6)=3/8
e) 3P3(1/6) (1/6) (1/6)=1/36
f) (3/6) (3/6) (3/6)=1/8
g) 3C1(3/6) (3/6) (3/6)=1/72
h) Two even numbers and one odd number or three odd numbers
=(3/6) (3/6) (3/6)+ (3/6) (3/6) (3/6)
=1/2
Q15
a) 4P4=24
b) The total sum of one digit
=(1+8+9+3)3P3
=126
The total sum of all formed number
=(126×100)＋(126×101)＋(126×102)＋(126×103)
=126+1260+12600+126000
=139986