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Mathematics: Post your doubts here!

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Hey everyone!
I have a question which is confusing me, can anyone answer that for me... explain me the steps to go thru.
In a group of 100 people, 40 own a cat, 25 own a dog and 15 own a cat and a dog. Find the probability that a person chosen at random:
(i) owns a dog or a cat [Answer: 0.5]
(ii) owns a dog or a cat but not both [Answer 0.35]
(iii) owns a dog given he owns a cat [Answer 0.375]
(iv) does not own a cat given that he owns a dog [Answer 0.4]

Thank you :)
 
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Hey everyone!
I have a question which is confusing me, can anyone answer that for me... explain me the steps to go thru.
In a group of 100 people, 40 own a cat, 25 own a dog and 15 own a cat and a dog. Find the probability that a person chosen at random:
(i) owns a dog or a cat [Answer: 0.5]
(ii) owns a dog or a cat but not both [Answer 0.35]
(iii) owns a dog given he owns a cat [Answer 0.375]
(iv) does not own a cat given that he owns a dog [Answer 0.4]

Thank you :)
in statistics "A or B" means "A or B or both"
P(A or B)=P(A)+P(B)-P(A and B)
P(A given B) = P(A|B)= P(A and B)/P(B)
so P(D)=0.25, P(C)=0.4, P(C and D)=0.15
i) P(D or C)=0.25+0.4-0.15=0.5
ii) P(D or C but not both)=0.5-0.15=0.35
iii)P(D|C)=P(D and C)/P(C)=0.15/0.4=0.375
iv) P(C`|D)=0.1/0.25=0.4 as 25 have dogs and 15 have both --> 10 have a dog but don't have a cat, so P(C` and D)=10/100=0.1
 
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Hi all. Can someone help me with this
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf
Q4(i) Why is the answer '45-50' instead of '45-49'? Because the < sign does not have equal sign
Q4(iv) Shouldn't we use Class boundaries in x-axis to draw the graph?
Mark scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_62.pdf

you have a cumulative frequency table, so the values given there are representing the upper bounds of each interval
q4i - the interval is 45 to 50 (45≤median<50)
q4iv - the class boundaries as used :)
 
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you have a cumulative frequency table, so the values given there are representing the upper bounds of each interval
q4i - the interval is 45 to 50 (45≤median<50)
q4iv - the class boundaries as used :)

I still don't really get it...
Another example: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
Q3(iii) Why is the midpoint calculated as 4.5 =(9+0)/2 and not 5 =(10+0)/2 ? Since the given info is already the boundaries, shouldn't we calculate midpoint from the boundaries?
Mark scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf

Another example: http://freeexampapers.automaticpapers.com/as-a-level/mathematics-9709/9709_s14_qp_62.pdf
Q6(iii)
Mark scheme: http://freeexampapers.automaticpapers.com/as-a-level/mathematics-9709/9709_s14_ms_62.pdf

The ways to get midpoints for these two questions are different.. How come?
 
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I still don't really get it...
Another example: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
Q3(iii) Why is the midpoint calculated as 4.5 =(9+0)/2 and not 5 =(10+0)/2 ? Since the given info is already the boundaries, shouldn't we calculate midpoint from the boundaries?
Mark scheme: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf

Another example: http://freeexampapers.automaticpapers.com/as-a-level/mathematics-9709/9709_s14_qp_62.pdf
Q6(iii)
Mark scheme: http://freeexampapers.automaticpapers.com/as-a-level/mathematics-9709/9709_s14_ms_62.pdf

The ways to get midpoints for these two questions are different.. How come?

for 9709_s14_qp_62
Time (seconds) is a type of continuous data --> the values are representing class boundaries
for 9709_s11_qp_63
Mark is a type of discrete data --> you should write the intervals: <10 means until and including 9, < 20 means until and including 19 etc so the intervals are 0≤mark≤9, 10≤mark≤19 etc and the class boundaries are -0.5 to 9.5, 9.5 to 19.5 etc
 
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in statistics "A or B" means "A or B or both"
P(A or B)=P(A)+P(B)-P(A and B)
P(A given B) = P(A|B)= P(A and B)/P(B)
so P(D)=0.25, P(C)=0.4, P(C and D)=0.15
i) P(D or C)=0.25+0.4-0.15=0.5
ii) P(D or C but not both)=0.5-0.15=0.35
iii)P(D|C)=P(D and C)/P(C)=0.15/0.4=0.375
iv) P(C`|D)=0.1/0.25=0.4 as 25 have dogs and 15 have both --> 10 have a dog but don't have a cat, so P(C` and D)=10/100=0.1
Thank you Maleko :D
 
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Hi. Can anyone help me with this S1 question please? Number 4(i)
 

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Hey guys I need help with Mechanics A2 question. May/June 2011 51. I don't understand the marking scheme. Please send a picture of the worked out answer. Thank you :)
 
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hey guys I'm new here, just joined few days ago, Nice to meet all of you :)
Can i get a little help on Question 6 (iii)? I don't quite get the marking scheme:( I really am bad in P & C:( Someone help me!:(
 

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Hey everyone!
I have a question which is confusing me, can anyone answer that for me... explain me the steps to go thru.
In a group of 100 people, 40 own a cat, 25 own a dog and 15 own a cat and a dog. Find the probability that a person chosen at random:
(i) owns a dog or a cat [Answer: 0.5]
(ii) owns a dog or a cat but not both [Answer 0.35]
(iii) owns a dog given he owns a cat [Answer 0.375]
(iv) does not own a cat given that he owns a dog [Answer 0.4]

Thank you :)
wat does the phrase ,within 1 standard deviation of the mean means????
 
Messages
283
Reaction score
506
Points
103
hey guys I'm new here, just joined few days ago, Nice to meet all of you :)
Can i get a little help on Question 6 (iii)? I don't quite get the marking scheme:( I really am bad in P & C:( Someone help me!:(
wat does the phrase within 1 stadard deviation of the mean ....,means??
 
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