Question

In: Statistics and Probability

smoker yes no no yes no no yes no no no yes no yes yes no...

smoker
yes
no
no
yes
no
no
yes
no
no
no
yes
no
yes
yes
no
no
yes
yes
no
yes
no
no
no
yes
yes
yes
no
yes
no
no
yes
no
yes
no
no
no
no
yes
no
yes
yes
no
no
no
no
yes
no

yes

lab scores before treatment
4,3
12,4
4,5
6,7
7,8
8,9
5,6
10,3
6,7
5,4
6,2
17,9
6,5
7,8
12,5
11,2
5,4
8,7
7,8
8,9
6,7
5,4
4,3
12,4
5,4
5,2
14,9
5,4
6,5
10,4
9,4
6,5
10,4
10,4
9,4
10,7
12,4
8,0
13,9
6,7
5,4
10,4
9,4
10,7
10,3
6,7
11,6

5,6

time since diagnosis
1
9
35
8
10
5
29
80
58
27
71
3
11
43
24
16
64
13
6
16
16
40
4
72
50
22
40
49
55
8
11
8
5
88
32
2
103
110
11
56
4
18
6
12
23
18
25
1

Question 3
Determine which distribution/distributions fit to the random variables: “X: smoking
status”, “Y: lab score before treatment” and “Z: time since diagnosis”. Explain all three
with reasons separately, and find their parameters. Provide the explanation.
So, implement the following steps for each variable:
(i: describe the random variable. ii: write down the name of distribution and the reason
you choose that distribution. iii: fill in the blanks, calculate the parameter/s)

3.a

i. X is the ......
ii. I expect ... distribution fits to X because ...
iii. X ~ .......(.....)
3.b

i.Y is the.....
ii. I expect .... distribution fits to Y because ....
iii. Y ~......(....)


3.c

i. Z is the.....
ii. I expect .... distribution fits to Z because ......
iii. Z ~......(....)

Solutions

Expert Solution

i. X is smoking status i.e. total no. of smokers.
ii. I expect Binomial distribution fits to X because each subject is either smoker or not, the somking status of a subject does not depend on other subjects, probability of smoking is fixed for all subjects.
iii. X ~ Binomial(48,20/48=0.4167).

3.b

i.Y is the lab score before treatment
ii. I expect normal distribution fits to Y because Y is continous variable and from QQ plot it is observed that almost all points falls closed to reference line.


iii. Y ~Normal(8.542, 3.067) where sample mean=8.542, sample sd=3.067.

3c. i. Z is time since diagnosis

ii. I expect exponential distribution fits to Z because Z is continuous random variable and from QQ plot it is clear that X follows exponential distribution:

\

(iii) Z follows exponential distribution with mean=29.54 (where sample mean=29.54).


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