# S3 helpWatch

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#1
[attach]797036[/attach

I need help with question 3

All I know is that the distribution is
N(u, 27040000/400)
1
8 months ago
#2
[attach]797036[/attach

I need help with question 3

All I know is that the distribution is
N(u, 27040000/400)
i) So what is the standard deviation for this distribution? Then, how does that relate to the £500 deviation?
1
#3
So would I equate 500 to k*(standard devition/square root of n) and try and find the value of k

I tried to do that earlier but it wasn't really working out
0
8 months ago
#4
So would I equate 500 to k*(standard devition/square root of n) and try and find the value of k

I tried to do that earlier but it wasn't really working out
What did you do, can you stick the numbers in?
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#5

this is what I did
0
8 months ago
#6

this is what I did
Looks good, use the cumulative tables to relate the 1.9 to a probability, remember its a two tailed test.

Edit - as a sanity check remember that 95% is about +/- 2 standard deviations which is about the value in this case.
Last edited by mqb2766; 8 months ago
0
#7

1.9 would give me 0.9713?
0
8 months ago
#8

1.9 would give me 0.9713?
Not quite right, but nearly.
.9726 - .5 = .4726
.4726*2 = 94.52%

Remember it starts in the middle (.5) so really you're looking at the area of the tail (0.0274), doubling it and subtracting from 1. Or subtract .5 and double. The table represents one tail.
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#9
Ohhhh yes I was supposed to use 1.923 not 1.9.

Just to clarify, the probability 0.9726 covers everything less than. But I want the middle so I subtract that probability from 1 to find the area of the tail multiply that by two, then subtract that new probability from 1 to get my answer.

Thank you for your responses, it makes so much sense now . I appreciate it a lot
0
8 months ago
#10
Ohhhh yes I was supposed to use 1.923 not 1.9.

Just to clarify, the probability 0.9726 covers everything less than. But I want the middle so I subtract that probability from 1 to find the area of the tail multiply that by two, then subtract that new probability from 1 to get my answer.

Thank you for your responses, it makes so much sense now . I appreciate it a lot
The 1.9 or 1.92 isn't too important (it is for marks, but not for the understanding).

As you say, the table represents the cumulative probability from -inf up to x. Its one tailed - the picture at the top shows this.
You want the similar distribution which is two tailed (symmetric about zero).
So find the "area" of the tail from the cumulative table, double it and subtract from one. That will transform the one tailed cumulative probability to a two tailed symmetric probability. Or, slightly quicker is to subtract .5 and double the result.

Also satisfies the sanity check: 2 standard deviations is ~95%.
0
#11
Hi I was wondering if you could help me with this question too

So I found out that the standard deviation of the differences is 1.838. To find the mean I need to find the differences between the dominant foot and the other foot add all those numbers up then divide that by 10 right?
My value for k would be 1.96
But for some reason I'm not getting the right answer

Attachment 797598
0
8 months ago
#12
Hi I was wondering if you could help me with this question too

So I found out that the standard deviation of the differences is 1.838. To find the mean I need to find the differences between the dominant foot and the other foot add all those numbers up then divide that by 10 right?
My value for k would be 1.96
But for some reason I'm not getting the right answer

Attachment 797598

Could you put the first part of the question in and can you put your working in for the mean and std dev? Saves me a job ...
0
#13

My lower limit for my confidence interval is correct but my upper limit isn't
0
8 months ago
#14

My lower limit for my confidence interval is correct but my upper limit isn't
I get the same mean, what is the confidence limits they give?
Is your std dev right - do they give that model solution?

Edit - scrub the query about part 1, I was misreading the question
Last edited by mqb2766; 8 months ago
0
#15
(Original post by mqb2766)
I get the same mean, what is the confidence limits they give?
Is your std dev right - do they give that model solution?
my answer to part i) is correct

But they got 1.91 as their upper limit
0
8 months ago
#16
my answer to part i) is correct

But they got 1.91 as their upper limit
I presume the 1.91 is a mistype for 3.91?

Four widths is
4 * 1.8/sqrt(10)
It can't be about 0.3 ...
0
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