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Thread starter 14 years ago
#1
It'd be amazing if you could help me out by doing these questions! Thanks!
0
14 years ago
#2
what exactly are you having trouble with ?
0
14 years ago
#3
well...what are you having trouble with? plotting the histogram is pretty simple, and you should be able to work out the mean, and therefore estimate the st.deviation...
those are the questions where the most marks are awarded..or is it the other stuff?
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Thread starter 14 years ago
#4
How do you do no.1? 0
14 years ago
#5
mean = sigma x / n
variance = sigma x^2 / n - mean^2
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Thread starter 14 years ago
#6
I got variance^2 = -64..so i cant work out the variance. 0
Thread starter 14 years ago
#7
Do you just use the positive root so the answer is 4
0
14 years ago
#8
(Original post by You)
Do you just use the positive root so the answer is 4
The question is wrong.
0
Thread starter 14 years ago
#9
Can someone do from 3b onwards plz
0
14 years ago
#10
(Original post by You)
Can someone do from 3b onwards plz
(b)
The observations are given in groups of different widths.

(c)
The median is defined to be the number m such that P(delay <= m) = 0.5.

P(delay <= 9.5) ~= (15 + 28 + 49)/200 = 0.46
P(delay <= 10.5) ~= (15 + 28 + 49 + 53)/200 = 0.725

To estimate the median we go a proportion (0.5 - 0.46)/(0.725 - 0.46) of the way from 9.5 to 10.5. That gives 9.651.

(d)
We represent each group of observations by its midpoint.

Delay | Number of motorists
------------------------------
05.0 | 15
07.5 | 28
09.0 | 49
10.0 | 53
11.5 | 30
14.0 | 15
18.0 | 10

sx
= (5*15 + 7.5*28 + 9*49 + 10*53 + 11.5*30 + 14*15 + 18*10)
= 1991

sxx
= (5^2*15 + 7.5^2*28 + 9^2*49 + 10^2*53 + 11.5^2*30 + 14^2*15 + 18^2*10)
= 21366.5

Estimate of mean = 1991/200 = 9.955
Estimate of variance = 21366.5/200 - 9.955^2 = 7.730
Estimate of standard deviation = sqrt(7.730) = 2.780

(e)
Estimate of coefficient of skewness
= 3(9.955 - 9.651)/2.780
= 0.328

(f)
For any normal distribution, the mean and the median are equal and so the coefficient of skewness is 0. The estimate in (e) is not close to zero. Hence the normal distribution may be a poor model for the data.
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