Experimental Design and Data Analysis

EXPERIMENTAL
DESIGN AND DATA
ANALYSIS
STUDENT ID:
[Pick the date]

Problem 1
(a) Graph
17 - 18 18 - 19 19 - 20 20- 21 21-22
0
0.2
0.4
0.6
0.8
1
1.2
CDF
(b) Mean of X
Mean = 19.05/1 = 19.05
(c) Median of X
Meadian=L+ ( n
2 −B )
G ∗W
Where,
2

Median would lie in the interval of 18-19.
L=lower class boundary of median group=18 ,
n=Total frequency=1
B=cumulative frequency of groupbefore median group=0.1 ,
G=Frequency of median group=0.45 ,
W =group width=1
Meadian=18+ ( 1
2 −0.1 )
0.45 ∗1=18.89
(d) Probability that age is more than 18 = 0.45+0.3+0.1+0.05 = 0.9
Probability of success = 0.9
Number of students = 5
Probability that at least four of them are over 18 (P)= P(4) + P(5)
Now,
Similarly,
3