Question

Answer part a and b of question. (I will give a negative rating if you solved just one part)

a) Find the 85th Percentile of the data set.

b) For a population represented by the sample data sets, suppose a vehicle is travelling at 40mph while the driver shifts his focus and fixates on a road object. Estimate a value for the distance traveled during the process that could be exceeded by this driver with a probability of 15%.

DATA SET 1:

Sample #	time sec
1	0.29
2	0.29
3	0.23
4	0.29
5	0.24
6	0.21
7	0.23
8	0.25
9	0.25
10	0.26
11	0.22
12	0.18
13	0.27
14	0.24
15	0.27
16	0.17
17	0.27
18	0.25
19	0.14
20	0.19
21	0.25
22	0.25
23	0.21
24	0.16
25	0.18
26	0.27
27	0.32
28	0.24
29	0.26
30	0.27

Data SET 2

Sample #	time sec
1	0.16
2	0.19
3	0.2
4	0.14
5	0.22
6	0.26
7	0.25
8	0.21
9	0.23
10	0.25
11	0.28
12	0.21
13	0.16
14	0.32
15	0.18
16	0.22
17	0.16
18	0.05
19	0.26
20	0.2
21	0.18
22	0.22
23	0.2
24	0.22
25	0.15
26	0.2
27	0.1
28	0.2
29	0.16
30	0.28

Answer 1

Here we'll be doing the calculation using EXCEL.
a)For finding the 0.85 percentile,PERCENTILE(array,k) function is used where "array" is the particular data set while "k" is the percentile needed.
so here using this function we get. 0.27 for set1 and 0.2565 for set 2.
b)Here we'll find the 0.15 percentile of the data which by using above function returns 0.1835s for set 1 and 0.16s for set 2
Hence the distance during this period =64*(5/18)*0.1835=3.262m for set 1

64*(5/18)*(0.16)=2.84m for set 2
Here we have used the conversions
1mile=1.6 KM => 40mile=64KM
and 1KM/H=(5/18)m/s

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