# S1 Edexcel June 2006 FeedbackWatch

12 years ago
#121

Venn diagram:
1/10
41/100
Can’t remember
2/3

Mean: 1.7 sth
Std. dev: 0.09…sth

New mean: 24.8

Can’t use the l/y/x formula for 90 degrees as it is outside the sample
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12 years ago
#122
(Original post by super_baros)
How did people get 1/25?.. I got 1/27..... Out of 55 phone calls, 11 were over 24.5 minutes.... So it two were taken out it'll be 11/55 * 10/54 = 1/27 right?....

And whoever asked about putting probabilities on Venn diagrams - you can, it just takes up more space and time. Doubt you'd lose marks for it.
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12 years ago
#123
How did people get 1/25?.. I got 1/27..... Out of 55 phone calls, 11 were over 24.5 minutes.... So it two were taken out it'll be 11/55 * 10/54 = 1/27 right?....
Thats exactly the same way i did it, but apparantly it's 2/5 = 0.4 not sure how people got that answer though maybe, 11/55 + 11/55 = 1/5 + 1/5 = 2/5?? ahhh i dunno bummer.
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12 years ago
#124
According to who? My stats teacher backed up the "sampling without replacement" method. Guess he might not have really been listening to us
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12 years ago
#125
According to who? My stats teacher backed up the "sampling without replacement" method. Guess he might not have really been listening to us
According to people in my class who total geniuses (genii?) who i'm to embarressed to question
*prays your stats teacher is right* that'll show the clever sods.
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12 years ago
#126
Hello people, i know everopne is busy but has any one uploaded todays S1 paper???or the answers

How did it go?
I found it bit wierd but ok

I thinked i flopped the last question becuase i didnt write down a percentage or fraction but just said how many people, i think thats wrong
hopefully they will give half marks
Do they give half marks?

Maths nearly over M1 to go
Yippeeeeee

doing-
Maths Further WHY
eco
english
Edit/Delete Message
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12 years ago
#127
(Original post by ~the_one~)
According to people in my class who total geniuses (genii?) who i'm to embarressed to question
*prays your stats teacher is right* that'll show the clever sods.
It'd make no sense for it to be 2/5, because it's quite clear from even looking at the data that 2/5 of the phone calls weren't over that length of time. You can't use addition - think of it as pullig coloured tokens out of a bag, and you'll see that multiplication wins.
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12 years ago
#128
This afternoon was my third attempt at S1 and i think the hardest exam (although that probably/hopefully means grade boundaries will be lower!)

I couldn't do the normal distribution question! :-( :-(

went as follows:

A high jumper can jump more than 1.68m for 7 in 10 jumps and more than 1.74cm for 1 in 5 jumps. Assuming the jump distribution to be modelled as the normal:

(a) Draw the normal distribution graph with the above information on [3]

(b) Calculate [mu] and [sigma] [6]

(c) find the probability that the jumper jumps more than 1.80 m on any given jump. [3]

I may have the wording wrong but the figures are (i think) correct.

Although I launched into this question with a good idea how to do it (namely section (b)) i soon came upon a problem! whilst it was obvious that both phi((1.68-[mu])/[sigma]) = 0.3 and phi((1.74-[mu])/[sigma]) = 0.8, i was unable to calculate a value for [mu] that was between 1.68 and 1.74 which it would have to be by definition, [mu] is the middle of the normal peak.

Can anyone help?

Other than that, I seem to have done well, mostly answers that have already been stated in other posts.
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12 years ago
#129
(Original post by Apagg)
It'd make no sense for it to be 2/5, because it's quite clear from even looking at the data that 2/5 of the phone calls weren't over that length of time. You can't use addition - think of it as pullig coloured tokens out of a bag, and you'll see that multiplication wins.
I remember putting down 1/5 for that Q...
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12 years ago
#130
I found S1 so difficult today - i got really confused on the normal distrubution question does anyone have any clue what the grade boundaries will be?
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12 years ago
#131
(Original post by x-Ro-x)
ok im really confused.... i thought that the grade boundaries never changed and were always the same - like 80%+ for an A, 70%-80% for a B...
I got 66/100 last january so whats that?
that's correct but the marks are scaled so as to make a fair representation of all grades... if it is a particularly difficult paper then the grade boundaries are lower but it will still be 80UMS for an A, 70-80UMS for a B etc. this is UMS though, not percentages - this is why maths papers are out of 75 but you get a mark from 100 - its scaled after every paper has been marked.

i got 66/100 too in january! lol thats a C though... if i get more than 71 today then i got my A :-)
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12 years ago
#132
(Original post by GεεRTHAN)
I remember putting down 1/5 for that Q...
As I recall, you had to work out the probability of two phone calls being over 24.5 minutes long. My friend put down 1/5 too, he just didn't read the question properly.
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12 years ago
#133
That phone call question was so random! come to think of it, isit even on the syllabus? i'm sure it's easy, but nothing's really easy unless you know the method of doing it. Weird, standard deviation wasn't even on there apart from the normal distibrution.
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12 years ago
#134
(Original post by kynastonpa)
This afternoon was my third attempt at S1 and i think the hardest exam (although that probably/hopefully means grade boundaries will be lower!)

I couldn't do the normal distribution question! :-( :-(

went as follows:

A high jumper can jump more than 1.68m for 7 in 10 jumps and more than 1.74cm for 1 in 5 jumps. Assuming the jump distribution to be modelled as the normal:

(a) Draw the normal distribution graph with the above information on [3]

(b) Calculate [mu] and [sigma] [6]

(c) find the probability that the jumper jumps more than 1.80 m on any given jump. [3]

I may have the wording wrong but the figures are (i think) correct.

Although I launched into this question with a good idea how to do it (namely section (b)) i soon came upon a problem! whilst it was obvious that both phi((1.68-[mu])/[sigma]) = 0.3 and phi((1.74-[mu])/[sigma]) = 0.8, i was unable to calculate a value for [mu] that was between 1.68 and 1.74 which it would have to be by definition, [mu] is the middle of the normal peak.

Can anyone help?

Other than that, I seem to have done well, mostly answers that have already been stated in other posts.
Values were actually 1.65 and 1.78, and I think less than 1.65 was 0.3, more than 1.78 was 0.2.
With those values

P(Z>(1.78-mu)/sigma) = 0.2

P(Z<(1.78-mu)/sigma)=0.8

Therefore (1.78-mu)/sigma = 0.8416 [Equation 1]

P(Z<(1.65-mu)/sigma) = 0.3
P(Z>(1.65-mu)/sigma) = 0.7

Therefore (1.65-mu)/sigma = -0.5244 [Equation 2]

1-2

0.13= 1.366 sigma
Sigma = 0.0951683748 (round as appropriate)

therefore mu = 1.6999906296

or 1.700 to 3 d.p.
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12 years ago
#135
correct
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12 years ago
#136
Do u get follow through marks? so if i figured out the wrong answers for normal distrubution in part B, will get 3 marks in part c if using my formula it was right???
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12 years ago
#137
is it okay to call the upper quartile the third quartile?
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12 years ago
#138
(Original post by Apagg)
Values were actually 1.65 and 1.78, and I think less than 1.65 was 0.3, more than 1.78 was 0.2.
With those values

P(Z>(1.78-mu)/sigma) = 0.2

P(Z<(1.78-mu)/sigma)=0.8

Therefore (1.78-mu)/sigma = 0.8416 [Equation 1]

P(Z<(1.65-mu)/sigma) = 0.3
P(Z>(1.65-mu)/sigma) = 0.7

Therefore (1.65-mu)/sigma = -0.5244 [Equation 2]

1-2

0.13= 1.366 sigma
Sigma = 0.0951683748 (round as appropriate)

therefore mu = 1.6999906296

or 1.700 to 3 d.p.
how do you get 0.8416 and -0.5244 to 4 decimal places. On my table it only had z values to 2 dp. and i got 0.77 (wrong) and -0.52 and the two z values. will i lose the marks for the others because i read the table wrong?
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12 years ago
#139
exactly - I only had 2 dp as well, so did our rest of us.
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12 years ago
#140
Do u get follow through marks? so if i figured out the wrong answers for normal distrubution in part B, will get 3 marks in part c if using my formula it was right???
yes you will. 95% certain of that
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