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The key points in the lecture slides of the Engineering Perspectives are:Normal Distribution, Gaussian Distribution, Continuous Probability, Density Function, Statistical Inference, Area Under Curve, Carbon Composition, Student’s T-Distribution, Standard Deviation, Confidence Interval
Typology: Slides
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1
− −
⋅
x
X
2 2
( )^2
2
(^1) σ
μ
σ π
Where π = 3.14159…, e = 2.71828…
(Eq. 1)
2
Graph of the Probability Dense
Function of Normal Distribution
σ 2 > σ 1
4
μ
Z =
Z
P( Z <1) = 0.
What is the total area under curve?
5
P( Z <- z ) = P ( Z > z ), for z >
Z =-1^ Z =
μ
P( Z <-1) = 0.159 (^) P( Z >1) = 0.
Z
7
Suppose the resistance of a carbon composition resistor is normally
distributed with mean μ = 1,000 and variance σ^2 = 900. What is the
probability that the resistor has measurement in excess of 1,060 ohms?
X ~ N ( μ =1000, σ=30), P ( X >1060)=?
X = the resistance of a carbon composition
Define X
8
df = ∞ (becomes normal distribution) df = 10 df = 5 df = 2 df = 1
population parameter to be estimated
10
“We are 95% confident that between 60% and 70% people will agree
with this proposal.”
95% is our confidence level; (60%, 70%) is our confidence interval
11
/ 2 / (^2) n
s
n
s
α α
where α is significance level and equal to (1 – confidence level)
t-distribution table can be downloaded from the course website
13
Sample Grades
1 88
2 85
3 90
4 87
5 85
6 80
7 75
8 92
9 95
10 85
75
2
=
=
s
x
Suppose these grades are sampled from a normal distribution with mean μ and standard deviation σ (unknown), find the confidence interval of the mean of the grades with 95% confidence level
t α/2 (10-1) = t α/2 (9) = 2.
The confidence interval for μ is: (86.2-2.262∙5.75/√10, 86.2+2.262∙5.75/√10) = (82.09, 91.31)
α = 0.05, n =
X = grades of student
14
−
x
X
λ λ (Eq. 7)
λ > 0 is a parameter of the distribution, often called the rate parameter
The cumulative distribution is given by
−
x
e x F x
λ x
Mean and Variance
E( X ) = 1/λ (Eq. 9) var( X ) = 1/λ^2 (Eq. 10)
16
Suppose the life in hours of a certain type of tube is a random variable that has an
exponential distribution with a mean of 1,000 hours. What is its probability density
function? What is the probability that such a tube will last at least 1,250 hours?
E( X ) = 1000 = 1/λ → λ = 1/
f ( x ) = 1/1000 × e(-1/1000)∙ x
P ( X ≥1250) = 1 - P ( X <1250) = 1- F ( X =1250) = 1- (1- e(-1/1000)∙1250^ ) = e(-1/1000)∙ = 0.
X = life in hours of the tube