Urrr... 9.45 is less than the mean is it not (I can't remember exactly that the mean for the distribution was)? If it is less than the mean, then it can't be the correct answer. P(X > mean) = 0.5, so P(X > mean - c) for c positive is going to be greater than 0.5, which does not allow the value given by the question of 0.2 (1 in 5).

Regards,

Reply 24

Unregistered

So what did you get?

Reply 25

Rich

Originally posted by Unregistered
So what did you get?

I can't remember exactly but I think it was 14. something or 15. something.

Does anyone remember what the mean and s.d. were for the distribution? Then I can work it out again and post the correct answer.

Regards,

mean=12
sd=3

For Z i got -0.85

Originally posted by Unregistered
mean=12
sd=3

So X follows N(12, 3^2). Z follows N(0, 1)

P(X > c) = 0.2
P(Z > (c - 12)/3) = 0.2
1 - P(Z < (c - 12)/3) = 0.2
P(Z < (c - 12)/3) = 0.8
So (c - 12)/3 = 0.84 (using big table for psi(z) = 0.7995 (closest))
Therefore c = 14.52

If you were to use the table over the page, which gives an exact value for 1 - psi(z) in our case, you can be more accurate and say:

P(X > c) = 0.2
P(Z > (c - 12)/3) = 0.2
(c - 12)/3 = 0.8416
Therefore c = 14.5248

The difference, as you can see, is almost negligible, and our teachers tell us that it doesn't matter which table you use.

Regards,

Reply 29

Unregistered

Now i c where i went wrong, I added in a minus!!

Reply 30

Surfing Hamster

Does anyone know how many marks the last question was worth in total, and which was the explanatory variable?

What was the total mark for the exam?

I didn't see question 7, and only did part (a) and a section of part (c) calculating B - how many marks will I pick up for this?

How many points do you need in the exam for an A/B/C/D etc?

Reply 31

Unregistered

i thought Q2 was hard. i did 1- probabilty of 0.8 so i got like 12.48,
I am pretty sure the explanatory variable was maths, since i read in my book explanatory = independant variable, i.e. variable that you change.
apart from that test was alright, too many graphs though i spent half the time drawing them. hope to get over 80%

Reply 32

Leekey

I got the same least squares regression line as some of you guys but my regression line was properly screwed !!!!
Any ideas ???(I think my y axis was M and my x axis was P if anyone did the same)

My answer for question 2 was 14.5(3.s.f) by the way.

Reply 33

Unregistered

Hello dudes. Just done the people just done the S1 paper again. Lifted a copy from school. These are the answers I got.

1.
Easy histogram (with class boundaries between i.e 7.5~8.5)
2.
I got t = 14.52 for life of battery
3a)
Sxy = -157.86
Sxx = 155.92
Syy = 214.96

3b)
PMCC= -0.862 to 3sf

3ci) PMCC unaffected by coding
cii) Strong negative correlation

4a) Proved k = 1/50
b) E(x) = 5.4
Var(x) = 0.52

4c) Var(2x-3) = 4Var(x) = 2.08

5 Dice
a) Uniform Distribution
b)E(x) = 3.5
Var(x) = 35/12
c) P(6,6,6) = 1/216
d)Score 16
6,6,4 6,4,6 4,6,6 5,5,6 5,6,5 6,5,5

So P(16) = 6/216 = 1/36

6a) Mean = 22 and 2/7
b) Unordered then ordered Stem & Leaf
c) Median = 20 , LQ = 13 , UQ = 31
d) IQR = 18 So no outliers
e) Box Whisker
f) Q2-Q1<Q3-Q2 so positive skew

7a) M explanatory
b) Scatter
c) Regression Line: 2.23 + 1.41x
d) Line best fit
e) Approx 23 in Physics if student got 15 in maths

Did everyone else get these or similar?:

Forgive me if there are any mistakes but I thought I would post somke solutions to put people out of their misery. Hope it went well dudes.

Reply 34

AnotherChris

Originally posted by Unregistered
Hello dudes. Just done the people just done the S1 paper again. Lifted a copy from school. These are the answers I got.

1.
Easy histogram (with class boundaries between i.e 7.5~8.5)
2.
I got t = 14.52 for life of battery
3a)
Sxy = -157.86
Sxx = 155.92
Syy = 214.96

3b)
PMCC= -0.862 to 3sf

3ci) PMCC unaffected by coding
cii) Strong negative correlation

4a) Proved k = 1/50
b) E(x) = 5.4
Var(x) = 0.52

4c) Var(2x-3) = 4Var(x) = 2.08

5 Dice
a) Uniform Distribution
b)E(x) = 3.5
Var(x) = 35/12
c) P(6,6,6) = 1/216
d)Score 16
6,6,4 6,4,6 4,6,6 5,5,6 5,6,5 6,5,5

So P(16) = 6/216 = 1/36

6a) Mean = 22 and 2/7
b) Unordered then ordered Stem & Leaf
c) Median = 20 , LQ = 13 , UQ = 31
d) IQR = 18 So no outliers
e) Box Whisker
f) Q2-Q1<Q3-Q2 so positive skew

7a) M explanatory
b) Scatter
c) Regression Line: 2.23 + 1.41x
d) Line best fit
e) Approx 23 in Physics if student got 15 in maths

Did everyone else get these or similar?:

Forgive me if there are any mistakes but I thought I would post somke solutions to put people out of their misery. Hope it went well dudes.

Fuck looking at it now, I did a number of mistakes.
-----------------------------------------------------------------
1) f) Q2-Q1<Q3-Q2 so positive skew

Here I put negative skew
----------------------------------------------------------------
2) I got t = 14.52 for life of battery
And this one I got wrong... Wasn't the question saying that 4/5 of batteries selected should have a higher average life...? Shouldn't that mean that you should get a value that translates to a negative value in the normal distribution, so that 4/5 of the area is after it?
-----------------------------------------------------------------
3) Score 16
6,6,4 6,4,6 4,6,6 5,5,6 5,6,5 6,5,5

So P(16) = 6/216 = 1/36
I only found 5 combinations here... :frown:

-----------------------------------------------------------------
4) The histogram I fucked up and did very hastily... I left it to the end, and I drew it wrong, one of the bars was 0.5 units wider (in fact I noticed it before the exam was over but since I didn't have enough time to finish, I actually wrote "this width is wrong by 0.5 units"
-----------------------------------------------------------------
5) Finally on the box plot, if the left wisker goes to below the mnimum value do you have to draw the whicker all the way to the end? For example here the lower boundary was -something, but the lowest value was 5... do we have to draw the whisker all the way to -something? Cause I only did it up to 5... doing bothe whiskers at the same length would pretty much ruin some of the predictive ability of the plot right? Or if I'm wrong I lost even more marks...

Well that's all my mistakes I think... Is there a chance I might miss the A?