![7/12/2012
General Continuous Distribution
b
Continuous Probability P[a<X<b] is the shaded area = ∫ f ( x) dx
a
Distributions f(.) is the probability density
function (pdf) of the continuous X
∞
Session 6 where f(x) ≥0 and
∫ f ( x)dx = 1
−∞
a b
∞ ∞
E ( X ) = (µ ) = ∫ xf ( x)dx; σ = ∫x f ( x)dx − µ 2
2 2
−∞ −∞
Note: P[X=c] = 0 for any c. 2
Cumulative Distribution Function
(CDF)
Numerical Exercise
• Find the mean, median, SD, IQR for a random variable
x with the following density function:
F ( x) = P[ X ≤ x] = ∫ f (t )dt
−∞
1/3
dF ( x) 1
exp ( −λ ( x − 3) )
f ( x) = 3
dx
1 3
3
Normal Distribution
N(µ,σ2)
(Cont.) Uniform distribution on (a,b)
• Symmetric, bell shaped, range -∞ to ∞
1
b−a
• mean = median= µ, standard deviation = σ,
skewness=kurtosis = 0
• Effective range µ - 3σ to µ + 3σ
a b • Fits many real data sets (specially large one)
• Z →N(0,1) standard normal distribution
Check: a + b 2 (b − a ) 2 • X → N(µ,σ2) implies (X- µ)/σ →N(0,1)
µ= ;σ =
2 12
5 6
1](https://image.slidesharecdn.com/session6-121004052559-phpapp01/85/Session-6-1-320.jpg)
![7/12/2012
Calculate Normal Probabilities
using tabulated values or Excel
P[0≤ Z≤ z] is given by the inside of the table (up to 4
decimal pts) for +ve z values
(upto 2 decimal pts) on the margin
Example:
0 z
Return from Stock A follows Normal Distribution
with mean 18% and standard deviation 8%. Find
P ( RA > 25%) P ( RA < 8%) P (10% < RA < 20%)
Reverse Calculation Clear-Tone Radio’s
• Return from Stock B follows Normal Distribution Let X be the no. of defective radios out of chosen 3.
and we know Let π = proportion of defectives in the box.
• Chance that it gives <5% return is only 2% Or equivalently D = No of defectives in the box
• Chance that it gives more than 25% return is 5%
Find X is random while π (or D) is not!
P ( RB > 25%) P( RB < 8%) P (10% < RB < 20%) Shankar is concerned with only this lot of 100 radios.
10
Clear-Tone Radio’s
What really matters is whether
D <= 10 (Good box) or D > 10 (bad box)
The decision about Box is good or bad has to taken on the basis of X
The possible options for Shankar is to reject the lot if and only if
A: X =3 B: X >= 2 C: X>=1
Clearly, in terms of the chance of accepting a bad box
option A would be worst while option C would be the best.
OTOH, in terms of the chance for rejecting Good box A > B>C.
To decide, one needs to de see which wrong decision is more critical.
Facts of the case suggests that rejecting a good box is more critical error.
So we need to ensure that the probability of accepting a box is kept at a
manageably low level.
Since, the chance of rejecting a Box increase with D, it is necessary and
enough to look at the distribution of X when D=10. 11
2](https://image.slidesharecdn.com/session6-121004052559-phpapp01/85/Session-6-2-320.jpg)
Session 6
- 1. 7/12/2012 General Continuous Distribution b Continuous Probability P[a<X<b] is the shaded area = ∫ f ( x) dx a Distributions f(.) is the probability density function (pdf) of the continuous X ∞ Session 6 where f(x) ≥0 and ∫ f ( x)dx = 1 −∞ a b ∞ ∞ E ( X ) = (µ ) = ∫ xf ( x)dx; σ = ∫x f ( x)dx − µ 2 2 2 −∞ −∞ Note: P[X=c] = 0 for any c. 2 Cumulative Distribution Function (CDF) Numerical Exercise • Find the mean, median, SD, IQR for a random variable x with the following density function: F ( x) = P[ X ≤ x] = ∫ f (t )dt −∞ 1/3 dF ( x) 1 exp ( −λ ( x − 3) ) f ( x) = 3 dx 1 3 3 Normal Distribution N(µ,σ2) (Cont.) Uniform distribution on (a,b) • Symmetric, bell shaped, range -∞ to ∞ 1 b−a • mean = median= µ, standard deviation = σ, skewness=kurtosis = 0 • Effective range µ - 3σ to µ + 3σ a b • Fits many real data sets (specially large one) • Z →N(0,1) standard normal distribution Check: a + b 2 (b − a ) 2 • X → N(µ,σ2) implies (X- µ)/σ →N(0,1) µ= ;σ = 2 12 5 6 1
- 2. 7/12/2012 Calculate Normal Probabilities using tabulated values or Excel P[0≤ Z≤ z] is given by the inside of the table (up to 4 decimal pts) for +ve z values (upto 2 decimal pts) on the margin Example: 0 z Return from Stock A follows Normal Distribution with mean 18% and standard deviation 8%. Find P ( RA > 25%) P ( RA < 8%) P (10% < RA < 20%) Reverse Calculation Clear-Tone Radio’s • Return from Stock B follows Normal Distribution Let X be the no. of defective radios out of chosen 3. and we know Let π = proportion of defectives in the box. • Chance that it gives <5% return is only 2% Or equivalently D = No of defectives in the box • Chance that it gives more than 25% return is 5% Find X is random while π (or D) is not! P ( RB > 25%) P( RB < 8%) P (10% < RB < 20%) Shankar is concerned with only this lot of 100 radios. 10 Clear-Tone Radio’s What really matters is whether D <= 10 (Good box) or D > 10 (bad box) The decision about Box is good or bad has to taken on the basis of X The possible options for Shankar is to reject the lot if and only if A: X =3 B: X >= 2 C: X>=1 Clearly, in terms of the chance of accepting a bad box option A would be worst while option C would be the best. OTOH, in terms of the chance for rejecting Good box A > B>C. To decide, one needs to de see which wrong decision is more critical. Facts of the case suggests that rejecting a good box is more critical error. So we need to ensure that the probability of accepting a box is kept at a manageably low level. Since, the chance of rejecting a Box increase with D, it is necessary and enough to look at the distribution of X when D=10. 11 2