STAT 2000 Lecture Notes - Lecture 13: Wii Fit, Random Variable
23 Feb 2017
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February 17, 2017
• Example 8: Wii Fit Study
o std. dev. 15 mean=111
▪
→ z-score
▪ percentile: enter mean=0, std. dev.=1;
o enter mean=111, std. dev.=15;
▪ 116.7799 corresponds to the 65th percentile
6.3 Probabilities When Observation Has Two Possible Outcomes
• 0= has type O blood; N=do not have type O blood
• P(O)=0.45; P(N)=0.55
Person 1
Person 2
Person 3
# w/ Type O
Blood
Probability
O
O
O
3
0.45*0.45*0.45=0.0911
O
O
N
2
0.45*0.45*0.55=
O
N
O
2
SAME
N
O
O
2
SAME
N
N
O
1
0.55*0.55*0.45=
N
O
N
1
SAME
O
N
N
1
SAME
N
N
N
0
0.55*0.55*0.55=0.1664
• 8 possibilities → 2*2*2=8
•
•
X
0
1
2
3
P(x)
0.1664
0.4084
0.3341
0.0911
• Using the binomial calculator
o n trials that has two possible outcomes → outcome of interest is called a success
and other outcome is called a failure
o each trial has saem probability of success denoted by p; probability of failure
denoted by q (1-p)
o each variable is independent
• Notation for binomial experiment
o n= # of trials
o p= probability of success for any given trials
o q= probability of failure for any given trial (q=1-p)
o x=# of successes for n trials
o n and p are the parameters of the binomial random variable
• Example 10: Drinking and Driving
o n=15 p=1/5=0.20
o
o
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