Question

MPG
36.3
41
36.9
37.1
44.9
36.8
30
37.2
42.1
36.7
32.7
37.3
41.2
36.6
32.9
36.5
33.2
37.4
37.5
33.6

1. The EPA collects data on 20 cars and calculates their gas mileage in miles per gallon (MPG).

d) Using the modified box-plot methodology determine if there are any outliers and justify. You do not have to make the box-plot!

e) Create a new variable by subtracting the mean from each observation and then dividing the difference by the standard deviation.

f) Find the mean, median and standard deviation of the new variable.

g) In one or two sentences, describe the original data.

Answer 1

We are given data

MPG
36.3
41
36.9
37.1
44.9
36.8
30
37.2
42.1
36.7
32.7
37.3
41.2
36.6
32.9
36.5
33.2
37.4
37.5
33.6

Let us first arrange the data in ascending order

30
32.7
32.9
33.2
33.6
36.3
36.5
36.6
36.7
36.8
36.9
37.1
37.2
37.3
37.4
37.5
41
41.2
42.1
44.9

There are a total of 20 observations

The second quartile = Median value of data, we can see that there are two central values which are 36.8 and 36.9

So, we take an average of both which is equal to 36.85.

The first quartile is the middle value of the lower half of data, here also two middle values which are 33.6 and 36.5 we take an average of both which is equal to 35.625

Now, similarly for the third quartile which is the middle value of the upper half of the data, so the middle values are 37.4 and 37.5 so, the average is equal to 37.425.

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Now the mean of the data =

Now, Variance =

Xi
30	-6.895	47.54103
32.7	-4.195	17.59803
32.9	-3.995	15.96003
33.2	-3.695	13.65303
33.6	-3.295	10.85703
36.3	-0.595	0.354025
36.5	-0.395	0.156025
36.6	-0.295	0.087025
36.7	-0.195	0.038025
36.8	-0.095	0.009025
36.9	0.005	0.000025
37.1	0.205	0.042025
37.2	0.305	0.093025
37.3	0.405	0.164025
37.4	0.505	0.255025
37.5	0.605	0.366025
41	4.105	16.85103
41.2	4.305	18.53303
42.1	5.205	27.09203
44.9	8.005	64.08002

Now,

Now, We know that Standard deviation is the square root of variance.

So, Standard deviation =

Xi
30	-6.895	-2.01694
32.7	-4.195	-1.22713
32.9	-3.995	-1.16862
33.2	-3.695	-1.08087
33.6	-3.295	-0.96386
36.3	-0.595	-0.17405
36.5	-0.395	-0.11555
36.6	-0.295	-0.08629
36.7	-0.195	-0.05704
36.8	-0.095	-0.02779
36.9	0.005	0.001463
37.1	0.205	0.059967
37.2	0.305	0.089219
37.3	0.405	0.118471
37.4	0.505	0.147724
37.5	0.605	0.176976
41	4.105	1.200802
41.2	4.305	1.259306
42.1	5.205	1.522576
44.9	8.005	2.341637

So, we have got new variables.

Now, Mean = -8.43769E-16

Similarly, Median = -0.01316

Xi
-2.01694	-2.01694	4.068038
-1.22713	-1.22713	1.505845
-1.16862	-1.16862	1.365683
-1.08087	-1.08087	1.168276
-0.96386	-0.96386	0.929025
-0.17405	-0.17405	0.030294
-0.11555	-0.11555	0.013351
-0.08629	-0.08629	0.007447
-0.05704	-0.05704	0.003254
-0.02779	-0.02779	0.000772
0.001463	0.001463	2.14E-06
0.059967	0.059967	0.003596
0.089219	0.089219	0.00796
0.118471	0.118471	0.014035
0.147724	0.147724	0.021822
0.176976	0.176976	0.03132
1.200802	1.200802	1.441925
1.259306	1.259306	1.585852
1.522576	1.522576	2.318238
2.341637	2.341637	5.483264

Variance =

Now, Standard deviation =

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We can clearly, see that on standarddizing the original data the standard deviation decreases to 1 from 3.44 which represents a consistent population values.

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