8/18/2019 Lab2 CEE 11
1/3
CEE 11 Lab #2 (Due in EEE Dropbox by 11:55pm Monday, Week 8)
Instruction: Toss a fair coin for 100 times.
H=head, T=tail
Q1. The number of heads is 54 ; the number of tails is
46_______
.
Q2. We define a random variable X as 1 if the outcome is head
and 0 if the outcome is tail. Then
your sample mean is 0.54 ; your sample variance is 0.2509 ;
your x = s2 =
sample standard deviation is 0.501 . Report numbers in Q1 and Q2
in the Google s =
spreadsheet (http://bit.ly/19nDwBB
).
Q3. If the coin is fair; i.e., if the probability of getting a
head is 0.5, what is the pmf of X, p(x)?What are its mean value, ,
and variance, ? What kind of random variable is it?μ X
σ
2 X
p(x)= 0.5, x = 1
0.5, x = 0
= x p(x) = 0.5(1) + 0.5(0) = 0.5μ X
Σ ∙
= = 1(0.5) + 0(0.5) - = 0.5 - 0.25 = 0.25σ2 X
[ x ( x)] μΣ 2 ∙ p
− X
2 .50 2
This is a Bernoulli Random Variable
Q4. If we denote the random sample mean by , then what are its
mean value, , and variance, X μ X
? What kind of random variable is it? Can you write down the pdf
of the random variable,σ2 X
( x) f
?
http://bit.ly/19nDwBBhttp://bit.ly/19nDwBB
8/18/2019 Lab2 CEE 11
2/3
= = 0.5μ X
μ X
= (1/n)
= (1/100)(0.25) = 0.0025σ2 X
σ2 X
is a normal random variable because n > 100. X
= f(x, , ) =( x) f μ σ
10.05√2π
∙ e 2(0.0025)−( x−0.5)2
Q5: What are the probabilities for to be greater than 0.45,
0.55, 0.65? X
p(
>0.45) = = = = 0.8413 X ( )1 −ϕ
σ
x−μ ( )1 −ϕ 0.050.45−0.5 (− )1 −ϕ 1
p(
>0.55) = = = = 0.1587 X ( )1 −ϕ
σ x−μ ( )1 −ϕ 0.05
0.55−0.5 (1)1 −ϕ
p( >0.65) = = = = 0.0013 X (
)1 −ϕ σ x−μ ( )1 −ϕ 0.05
0.65−0.5 (3)1 −ϕ
Q6: Use the data on the google spreadsheet
(http://bit.ly/19nDwBB
) to create a histogram of
sample means. Also plot the pdf in Q4 on the same picture and
see if it makes sense or not.
Page 2 of 3
http://bit.ly/19nDwBB
