• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

How did STATISTICS 1 go?

What do you think?

  • Total voters
    98
  • Poll closed .
R

randomcod

Messages
188
Reaction score
204
Points
53
thementor said:
seriously man -_-
Click to expand...
It is 60. I accidentally multiplied by 5 (please don't ask why, so many 5P something got the 5 stuck in my head and I subsconciously wrote down 5 instead of 3 and got 100), but yes, the answer is 60. It was actually one of the easier questions IMO. It had to be 60 because the paper clearly stated NO REPETITIONS for this question. 5P2 for each x 3 Good luck to everyone.
 
K

Kamihus

Messages
930
Reaction score
2,173
Points
253
The mid-points and mean have been updated. The ones mentioned previously were incorrect.

Q1 [4 marks]
n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3
Ans: 0.813

Q2 [4 marks]
Group A was of 7, Group B was of 2, Group C was of 2. Choose 5 and at least 1 from each group.
Possibilities:
3 from A and 1 from B and 1 from C. 7C3*2C1*2C1=140
2 from A and 2 from B and 1 from C. 7C2*2C2*2C1=42
2 from A and 1 from B and 2 from C. 7C2*2C1*2C2-42
1 from A and 2 from B and 2 from C. 7C1*2C2*2C2=7
Add them.
Ans: 231

Q3 (i) [3 marks] (ii) [2 marks]
Roger had a probability of winning first set=0.6
Roger had a probability of winning second set if he had won the first set=0.7
Roger had a probability of winning second set if he lost the first set=0.25
Match would end when any player wins 2 sets and match cannot be drawn and Andy was the other player.

(i) Probability that match would end in 2 sets.
If Roger wins: 0.6*0.7=0.42
If Andy wins: 0.4*0.75=0.3
Add them.

(ii) If match ends in 2 sets find the probability that match Andy won the match
0.3/0.72
Ans: (i) 0.72 (ii) 0.417

Q4 (i) [3 marks] (ii) [4 marks] (iii) [2 marks]
Coin A is thrown twice and Coin B once
Coin A has probability of head= 2/3
Coin B has probability of head= 1/4

(i) Show that the probability of getting 1 head is 13/36
If Coin A gives 1 head and Coin B gives none: 2*2/3*1/3*3/4=12/36 (Multiplied by 2 as both throws on Coin A can give a head)
If Coin A gives 0 heads and Coin B gives 1 head: 1/3*1/3*1/4=1/36
Add them and total= 13/36

(ii) Draw a probability distribution table of numbers of heads obtained
0 when no heads obtained: 1/3*1/3*3/4=3/36
2 when Coin A gives 2 heads: 2/3*2/3*3/4=12/36
Both Coin A and Coin B give one head each: 2*1/3*2/3*1/4=4/36 (Multiplied by 2 as both throws on Coin A can give a head)
Add them.
3 when all heads: 2/3*2/3:1/4=4/36

(iii) Find the expected value
1*13/36 + 2*16/36 +3*4/36=19/12
Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Q5 (i) [2 marks] (ii) [3 marks] (iii) [3 marks]
1345789 is a number

(i) Find the number of arrangements when odd numbers together and digits not repeated
5!*3! =720

(ii) Arrangements between 3000 and 5000 and even numbers and digits not repeated
When starting with 3: Last digit can be 4 or 8 so the middle two digits are 5P2=5P2*2!=40
When starting with 4: Last digit is 8 so middle two digits are 5P2=20
Total arrangements = 60

(iii) Multiples of 5 less than 1000 and digits ARE REPEATED
Last digit has to be 5 and can be 3-digit, 2-digit or 1-digit number
When 3 digit number: 7*7*1=49
When 2 digit number: 7*1=7
When 1-digit number: 1
Total =57
Ans: (i) 720 (ii) 60 (iii) 57

Q6 (i) [1 mark] (ii) [4 marks] (iii) [4 marks]
Total athletes=57
<10 were 0 <10.5 were 4 <11.0 were 10 <12.0 were 40 <12.5 were 49 <13.5 were 57

(i) Find in the interval 10.5-11.0 =6

(ii) Draw a histogram
Frequencies: 4,6,30,9,8
Frequency densities: 8,12,30,18,8

(iii) Calculate mean and variance
Mid-points: 10.25, 10.75, 11.5, 12.25, 13.0
Mean=11.7
Variance=0.547
Ans: (i) 6 (iii) 11.7, 0.547

Q7 (i) [3 marks] (ii) [2 marks] (iii) [6 marks]
Rafa studies at mean rate 1.9. The probability of studying more than 1.35 is 0.8

(i) Find Standard Deviation
Z=0.842 so S.D. =0.653

(ii) Find the probability that studies less than 2.0
(2.0-1.9)/0.653=0.1531
P=0.561

(iii) 200 samples are taken. Find the probability between 163 and 173 inclusive
np=160 npq=32
Less than 173.5: 0.9915
Less than 162.5: 0.6707
Between them= 0.321
Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes :)
 
Last edited:
talha196

talha196

Messages
118
Reaction score
62
Points
38
randomcod said:
It is 60. I accidentally multiplied by 5 (please don't ask why, so many 5P something got the 5 stuck in my head and I subsconciously wrote down 5 instead of 3 and got 100), but yes, the answer is 60. It was actually one of the easier questions IMO. It had to be 60 because the paper clearly stated NO REPETITIONS for this question. 5P2 for each x 3 Good luck to everyone.
Click to expand...
lol dont worry man you know i made the silliest mistake on cud ever make in the third part... it was repition allowed and i did it right on the question paper but god damn when i did it on answer sheet i forgot that repitition is allowed.... facepalm,
 
thementor

thementor

Messages
943
Reaction score
805
Points
103
Kamihus said:
Q1 [4 marks]
n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3
Ans: 0.813

Q2 [4 marks]
Group A was of 7, Group B was of 2, Group C was of 2. Choose 5 and at least 1 from each group.
Possibilities:
3 from A and 1 from B and 1 from C. 7C3*2C1*2C1=140
2 from A and 2 from B and 1 from C. 7C2*2C2*2C1=42
2 from A and 1 from B and 2 from C. 7C2*2C1*2C2-42
1 from A and 2 from B and 2 from C. 7C1*2C2*2C2=7
Add them.
Ans: 231

Q3 (i) [3 marks] (ii) [2 marks]
Roger had a probability of winning first set=0.6
Roger had a probability of winning second set if he had won the first set=0.7
Roger had a probability of winning second set if he lost the first set=0.25
Match would end when any player wins 2 sets and match cannot be drawn and Andy was the other player.

(i) Probability that match would end in 2 sets.
If Roger wins: 0.6*0.7=0.42
If Andy wins: 0.4*0.75=0.3
Add them.

(ii) If match ends in 2 sets find the probability that match Andy won the match
0.3/0.72
Ans: (i) 0.72 (ii) 0.417

Q4 (i) [3 marks] (ii) [4 marks] (iii) [2 marks]
Coin A is thrown twice and Coin B once
Coin A has probability of head= 2/3
Coin B has probability of head= 1/4

(i) Show that the probability of getting 1 head is 13/36
If Coin A gives 1 head and Coin B gives none: 2*2/3*1/3*3/4=12/36 (Multiplied by 2 as both throws on Coin A can give a head)
If Coin A gives 0 heads and Coin B gives 1 head: 1/3*1/3*1/4=1/36
Add them and total= 13/36

(ii) Draw a probability distribution table of numbers of heads obtained
0 when no heads obtained: 1/3*1/3*3/4=3/36
2 when Coin A gives 2 heads: 2/3*2/3*3/4=12/36
Both Coin A and Coin B give one head each: 2*1/3*2/3*1/4=4/36 (Multiplied by 2 as both throws on Coin A can give a head)
Add them.
3 when all heads: 2/3*2/3:1/4=4/36

(iii) Find the expected value
1*13/36 + 2*16/36 +3*4/36=19/12
Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Q5 (i) [2 marks] (ii) [3 marks] (iii) [3 marks]
1345789 is a number

(i) Find the number of arrangements when odd numbers together and digits not repeated
5!*3! =720

(ii) Arrangements between 3000 and 5000 and even numbers and digits not repeated
When starting with 3: Last digit can be 4 or 8 so the middle two digits are 5P2=5P2*2!=40
When starting with 4: Last digit is 4 so middle two digits are 5P2=20
Total arrangements = 60

(iii) Multiples of 5 less than 1000 and digits ARE REPEATED
Last digit has to be 5 and can be 3-digit, 2-digit or 1-digit number
When 3 digit number: 7*7*1=49
When 2 digit number: 7*1=7
When 1-digit number: 1
Total =57
Ans: (i) 720 (ii) 60 (iii) 57

Q6 (i) [1 mark] (ii) [4 marks] (iii) [4 marks]
Total athletes=57
<10 were 0 <10.5 were 4 <11.0 were 10 <12.0 were 40 <12.5 were 49 <13.5 were 57

(i) Find in the interval 10.5-11.0 =6

(ii) Draw a histogram
Frequencies: 4,6,30,9,8
Frequency densities: 8,12,30,18,8

(iii) Calculate mean and variance
Mid-points: 10.2, 10.7, 11.45, 12.2, 12.95
Mean=11.6
Variance=0.547
Ans: (i) 6 (iii) 11.6, 0.547
Q7 (i) [3 marks] (ii) [2 marks] (iii) [6 marks]
Rafa studies at mean rate 1.9. The probability of studying more than 1.35 is 0.8

(i) Find Standard Deviation
Z=0.842 so S.D. =0.653
(ii) Find the probability that studies less than 2.0
(2.0-1.9)/0.653=0.1531
P=0.561

(iii) 200 samples are taken. Find the probability between 163 and 173 inclusive
np=160 npq=32
Less than 173.5: 0.9915
Less than 162.5: 0.6707
Between them= 0.321
Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes :)
Click to expand...
Q3(II) is wrong. The answer is 0.583
The conditional probability will be (0.6*0.7)/0.72 = 0.583
 
I

IGCSE13

Messages
73
Reaction score
133
Points
18
Ho
Kamihus said:
Q1 [4 marks]
n=19 probability of success=0.12, find fewer than 4 so apply binomial with 0,1,2 and 3
Ans: 0.813

Q2 [4 marks]
Group A was of 7, Group B was of 2, Group C was of 2. Choose 5 and at least 1 from each group.
Possibilities:
3 from A and 1 from B and 1 from C. 7C3*2C1*2C1=140
2 from A and 2 from B and 1 from C. 7C2*2C2*2C1=42
2 from A and 1 from B and 2 from C. 7C2*2C1*2C2-42
1 from A and 2 from B and 2 from C. 7C1*2C2*2C2=7
Add them.
Ans: 231

Q3 (i) [3 marks] (ii) [2 marks]
Roger had a probability of winning first set=0.6
Roger had a probability of winning second set if he had won the first set=0.7
Roger had a probability of winning second set if he lost the first set=0.25
Match would end when any player wins 2 sets and match cannot be drawn and Andy was the other player.

(i) Probability that match would end in 2 sets.
If Roger wins: 0.6*0.7=0.42
If Andy wins: 0.4*0.75=0.3
Add them.

(ii) If match ends in 2 sets find the probability that match Andy won the match
0.3/0.72
Ans: (i) 0.72 (ii) 0.417

Q4 (i) [3 marks] (ii) [4 marks] (iii) [2 marks]
Coin A is thrown twice and Coin B once
Coin A has probability of head= 2/3
Coin B has probability of head= 1/4

(i) Show that the probability of getting 1 head is 13/36
If Coin A gives 1 head and Coin B gives none: 2*2/3*1/3*3/4=12/36 (Multiplied by 2 as both throws on Coin A can give a head)
If Coin A gives 0 heads and Coin B gives 1 head: 1/3*1/3*1/4=1/36
Add them and total= 13/36

(ii) Draw a probability distribution table of numbers of heads obtained
0 when no heads obtained: 1/3*1/3*3/4=3/36
2 when Coin A gives 2 heads: 2/3*2/3*3/4=12/36
Both Coin A and Coin B give one head each: 2*1/3*2/3*1/4=4/36 (Multiplied by 2 as both throws on Coin A can give a head)
Add them.
3 when all heads: 2/3*2/3:1/4=4/36

(iii) Find the expected value
1*13/36 + 2*16/36 +3*4/36=19/12
Ans : (ii) 0: 3/36 1:13/36 2: 16/36 3:4/36 (iii) 1.58

Q5 (i) [2 marks] (ii) [3 marks] (iii) [3 marks]
1345789 is a number

(i) Find the number of arrangements when odd numbers together and digits not repeated
5!*3! =720

(ii) Arrangements between 3000 and 5000 and even numbers and digits not repeated
When starting with 3: Last digit can be 4 or 8 so the middle two digits are 5P2=5P2*2!=40
When starting with 4: Last digit is 4 so middle two digits are 5P2=20
Total arrangements = 60

(iii) Multiples of 5 less than 1000 and digits ARE REPEATED
Last digit has to be 5 and can be 3-digit, 2-digit or 1-digit number
When 3 digit number: 7*7*1=49
When 2 digit number: 7*1=7
When 1-digit number: 1
Total =57
Ans: (i) 720 (ii) 60 (iii) 57

Q6 (i) [1 mark] (ii) [4 marks] (iii) [4 marks]
Total athletes=57
<10 were 0 <10.5 were 4 <11.0 were 10 <12.0 were 40 <12.5 were 49 <13.5 were 57

(i) Find in the interval 10.5-11.0 =6

(ii) Draw a histogram
Frequencies: 4,6,30,9,8
Frequency densities: 8,12,30,18,8

(iii) Calculate mean and variance
Mid-points: 10.2, 10.7, 11.45, 12.2, 12.95
Mean=11.6
Variance=0.547
Ans: (i) 6 (iii) 11.6, 0.547
Q7 (i) [3 marks] (ii) [2 marks] (iii) [6 marks]
Rafa studies at mean rate 1.9. The probability of studying more than 1.35 is 0.8

(i) Find Standard Deviation
Z=0.842 so S.D. =0.653
(ii) Find the probability that studies less than 2.0
(2.0-1.9)/0.653=0.1531
P=0.561

(iii) 200 samples are taken. Find the probability between 163 and 173 inclusive
np=160 npq=32
Less than 173.5: 0.9915
Less than 162.5: 0.6707
Between them= 0.321
Ans: (i) 0.653 (ii) 0.561 (iii) 0.321

Feel free to post any queries or mistakes :)[/
Click to expand...
Which formula did you use for the mean and variance in q6iii
 
You must log in or register to reply here.
Top