Edexcel (IAL): Statistics 1, WST01 (27th January 2016)

Original post by Awaiskorai

Either we have all messed up or are we actually getting a 90+ UMS? XD Not sure.

I'm not sure what you're trying to say.

Reply 62

Awaiskorai

I am trying to say, as we have got almost the same answers, either we have fared great or everything is wrong and down the hill xP

About the negative sales part I thought it could be negative because of a loss in sales :/ I would have certainly rechecked if I had time.

Reply 63

Original post by Zacken

Disclaimer: If you disagree with any of my answers, I'm probably wrong and you're probably right

1. (a) F(5) = 5/6
(b) E(X) = 2
(c) Var(X) = 15/2
(d) (i) E(Y) = 3
(ii) Var(Y) = 30
(iii) P(Y > X) = 5/12

2. (a) 30 marks is the lowest pass mark.
(b) 46 marks is the lowest merit mark.
(c) c=70, d=6
(d) 68, 72, 79 are the three highest marks
(e) Outlier at 5, range starts at 10, not sure why they mentioned 15, did I do something wrong?
(f) 0.5 * 0.25 * 0.25 * 3 = 3/32

I hate box plots, probably failed that question.

3. (a)S_vs = 43.911 and S_ss = 40.44
(b) pmcc = 0.929
(c) pmcss approx 1 blah blah blah
(d) s = 1.72 + 0.794 x
(e) y = 1920 + 3.97x
(f) coeff of x higher for textbooks blah blah blah

4. (a) 0.3, 0.5, 0.7, 0.9 - those are the fails. Success on the last is 0.1
(b) 0.9055 = 0.7 + 0.3 * 0.5 + 0.3 * 0.5 * 0.3 + 0.3 * 0.5 * 0.7 * 0.1
(c) 1700/1811
(d)Shall we make a petition? Hannah's sweet much. p + (1-p)(p-0.2) = 0.95
(e) 1.1 - sqrt(3/50) approx 0.855

5(a) 0.3446
(b) 0.0583
(c) mean = 22.57, standard dev = 5.43

6(a) mean = 836, standard dev = 79.9
(b) postiive skew, mean > median.
(c) no change
(d) standard dev decrease.

For 2(f) I used the P(A|B) formula. 'A' being the probability of two of them getting merit, and 'B' being the probability of three of them passing. Do you think this would work..?

Reply 64

Ayman!

15

Original post by Zacken

6(d) quantitatively? You get something like sqrt(5000 something), don't you? Lower value is correct, I said it decreases? Although my reason was:

60^2 + 60^2 = 7200 < 2 \times \frac{68 430}{10}

or whatever the 68430 was. We should both get the marks.

Thank god for the 2(e), why did they mention 15 marks, though?

How'd you do 2(f) and what did you do that differs from mine, what did I mess up? :eek:

Congratulations! This was a 6-question paper, those generally have lower grade boundaries.

Yep, I think I got 73.9. I noticed that

Unparseable latex formula:

\[\sum (x-\bar{x})^{2} = S_{xx}

, so I just found the new

\sum x^{2}

by finding

Unparseable latex formula:

\frac{old \sum x^{2} + (value_{1})^{2} + (value_{2})^{2}}{12}\]

. After that, it was just the regular formula.

It was just to trip us up. I realised the moment I read it. :tongue:

For 2(f) what I did was 3 x (0.75)²x (0.25), since 0.75 students pass and 0.25 students get a merit. I might be wrong though.

The normal distribution question was really nice. Plenty of people at my centre couldn't do it from the general reaction. It's definitely a boundary dropper! (which makes it nicer for me 'cause I could do it :colone:

)

See you on the M2 thread. :wink:

EDIT: Apologising for the terrible latex :frown:

Reply 65

Awaiskorai

How many marks are we going to lose for negative intercept? Is Ecf there?
And I looked at the normal distribution table.

Reply 66

Awaiskorai

Original post by aymanzayedmannan

Yep, I think I got 73.9. I noticed that

Unparseable latex formula:

\[\sum (x-\bar{x})^{2} = S_{xx}

, so I just found the new

\sum x^{2}

by finding

Unparseable latex formula:

\frac{old \sum x^{2} + (value_{1})^{2} + (value_{2})^{2}}{12}\]

. After that, it was just the regular formula.

It was just to trip us up. I realised the moment I read it. :tongue:

For 2(f) what I did was 3 x (0.75)²x (0.25), since 0.75 students pass and 0.25 students get a merit. I might be wrong though.

See you on the M2 thread. :wink:

EDIT: Apologising for the terrible latex :frown:

That's certainly wrong because two students that passed were on merit so (.25*.25 *. 75) and there were three such possibilities so, 3(.25*.25*.75)

Reply 67

Ayman!

15

Original post by Awaiskorai

That's certainly wrong because two students that passed were on merit so (.25*.25 *. 75) and there were three such possibilities so, 3(.25*.25*.75)

Did you get a final answer of 9/64? I got that.

Reply 68

Original post by Awaiskorai

I am trying to say, as we have got almost the same answers, either we have fared great or everything is wrong and down the hill xP

I doubt we all got the same wrong answer. :tongue:

About the negative sales part I thought it could be negative because of a loss in sales :/ I would have certainly rechecked if I had time.

That doesn't make sense, oh well - you shan't lose too many marks. Maybe 1 or 2.

Original post by elhm18

For 2(f) I used the P(A|B) formula. 'A' being the probability of two of them getting merit, and 'B' being the probability of three of them passing. Do you think this would work..?

I'll try your method and get back to you in a few minutes.

Original post by Awaiskorai

And I looked at the normal distribution table.

You were meant to look at the percentage points table which give more exact values for z.

Original post by aymanzayedmannan

Yep, I think I got 73.9. I noticed that

Unparseable latex formula:

\[\sum (x-\bar{x})^{2} = S_{xx}

, so I just found the new

\sum x^{2}

by finding

Unparseable latex formula:

\frac{old \sum x^{2} + (value_{1})^{2} + (value_{2})^{2}}{12}\]

. After that, it was just the regular formula.

Sounds right. I did that as well but I was wary of using nothing but calculations so I explained it the other way just to be safe.

It was just to trip us up. I realised the moment I read it. :tongue:

For 2(f) what I did was 3 x (0.75)²x (0.25), since 0.75 students pass and 0.25 students get a merit. I might be wrong though.

Phew! I'm not very good with box plots, so I stayed confused. :tongue:

I think you mean 3 x (0.25)^2 x (0.75) - I considered doing that as well, but I figured that the 0.75 should be a 0.5 because otherwise it includes people getting merits. In either case, neither of us should lose more than 2 marks.

The normal distribution question was really nice. Plenty of people at my centre couldn't do it from the general reaction. It's definitely a boundary dropper! (which makes it nicer for me 'cause I could do it :colone:

)

It was very nice! The mark distribution was nice as well - 3, 5, 6. Looks like you're on track for full UMS. :colone:

See you on the M2 thread. :wink:

Now that isn't going to go well...

EDIT: Apologising for the terrible latex :frown:

Don't apologise! :smile:

Reply 69

Fuwad98

1

Original post by Zacken

Disclaimer: If you disagree with any of my answers, I'm probably wrong and you're probably right

1. (a) F(5) = 5/6
(b) E(X) = 2
(c) Var(X) = 15/2
(d) (i) E(Y) = 3
(ii) Var(Y) = 30
(iii) P(Y > X) = 5/12

2. (a) 30 marks is the lowest pass mark.
(b) 46 marks is the lowest merit mark.
(c) c=70, d=6
(d) 68, 72, 79 are the three highest marks
(e) Outlier at 5, range starts at 10, not sure why they mentioned 15, did I do something wrong?
(f) 0.5 * 0.25 * 0.25 * 3 = 3/32

I hate box plots, probably failed that question.

3. (a)S_vs = 43.911 and S_ss = 40.44
(b) pmcc = 0.929
(c) pmcss approx 1 blah blah blah
(d) s = 1.72 + 0.794 x
(e) y = 1920 + 3.97x
(f) coeff of x higher for textbooks blah blah blah

4. (a) 0.3, 0.5, 0.7, 0.9 - those are the fails. Success on the last is 0.1
(b) 0.9055 = 0.7 + 0.3 * 0.5 + 0.3 * 0.5 * 0.3 + 0.3 * 0.5 * 0.7 * 0.1
(c) 1700/1811
(d)Shall we make a petition? Hannah's sweet much. p + (1-p)(p-0.2) = 0.95
(e) 1.1 - sqrt(3/50) approx 0.855

5(a) 0.3446
(b) 0.0583
(c) mean = 22.57, standard dev = 5.43

6(a) mean = 836, standard dev = 79.9
(b) postiive skew, mean > median.
(c) no change
(d) standard dev decrease.

All right except 2(f) if I am not wrong...

Reply 70

Original post by Awaiskorai

That's certainly wrong because two students that passed were on merit so (.25*.25 *. 75) and there were three such possibilities so, 3(.25*.25*.75)

But 0.75 is the probability of getting a merit or passing. 0.5 is the probability of passing only.

Reply 71

Original post by Fuwad98

All right except 2(f) if I am not wrong...

Lots of people seem to be saying that my 2(f) is wrong, it probably is.

I'm not convinced why you do 3 * 0.25 * 0.25 * 0.75 though, surely the 0.75 should be (0.75 - 0.25) = 0.5 instead.

Reply 72

Original post by Zacken

Lots of people seem to be saying that my 2(f) is wrong, it probably is.

I'm not convinced why you do 3 * 0.25 * 0.25 * 0.75 though, surely the 0.75 should be (0.75 - 0.25) = 0.5 instead.

I think getting a merit instantly means you'd be passing. They intersect. And 75% do pass, even if it includes the ones who got a merit.

That's what I think, at least.

Reply 73

Original post by elhm18

I think getting a merit instantly means you'd be passing. They intersect. And 75% do pass, even if it includes the ones who got merit.

That's what I think, at least.

But the question specifically said only 2 get a merit and 1 gets a pass.

Reply 74

Original post by Zacken

But the question specifically said only 2 get a merit and 1 gets a pass.

Did it? I remember it saying "given three of the students pass, what is the probability of two getting a merit?" or something along the lines. I'm not entirely sure, but that's what I remember.

Reply 75

Original post by elhm18

Did it? I remember it saying "given three of the students pass, what is the probability of two getting a merit?" or something along the lines. I'm not entirely sure, but that's what I remember.

I'm very sure it used the world "only" specifically, had it not - I'd have done the same thing as you.

Edit: anyways, I'm not fussed. Either one would get 1/2 marks and that's fine so I'm going to stop arguing now. The boundaries should be fairly low anyway, with that normal dist. quesiton.

(edited 8 years ago)

Reply 76

Original post by Zacken

I'm very sure it used the world "only" specifically, had it not - I'd have done the same thing as you.

Hm, maybe so. I might have read wrong, but I still feel as though they mentioned, "given three pass"

I would have not used the P(A|B) formula otherwise.

We'll get to know when the time comes, haha. But you're probably right. :tongue:

Reply 77

Original post by elhm18

Hm, maybe so. I might have read wrong, but I still feel as though they mentioned, "given three pass"

I would have not used the P(A|B) formula otherwise.

We'll get to know when the time comes, haha. :tongue:

Sooner, somebody will put up model answers within the next few days. :smile:

Reply 78