ch04 - part 2 (Cont unifm dist).ppt explained

Good lecture and detailed

1. Continuous Probability
Distribution
* Uniform Distribution
* Normal Distribution

2. Definition
4-5 Continuous Uniform Distribution
-Discrete uniform distribution, f(x) = 1/n

3. Figure 4-8Continuous uniform probability density
function.
4-5 Continuous Uniform Distribution
Discrete uni. dis

4. Mean and Variance
4-5 Continuous Uniform Distribution

5. Example
4-5 Continuous Uniform Distribution
To know that it is unifm. dist
f(x)=1/ (b-a)
=1/ (20-0) =0.05

6. Figure 4-9Probability for Example 4-9.
4-5 Continuous Uniform Distribution

7. 4-5 Continuous Uniform Distribution

8. Text Book Problem
•441
a.F(X) = (x-0.9)/0.15
b.P(X>1.02) = 0.2
c.Here you have to identify the value of x such
that the function will equate to 0.9 or 90%
i.e., P(X>x)=0.9
x=0.915
d. E(X)=0.976; V(X)=0.00188

9. Text Book Problem
•446 (Please note that there is a mistake in
the problem. Change 746 with 374 and 752
with 380)
a.E(X) = 377; s(X)=1.732
b.P(X<375) = 0.1667
c.P(X>x)=0.95; x=374.3
d.Mean extra cost = 0.004/ container
(Hint: mean extra cost = cost* mean extra
volume)

10. Normal
Distribution

11. •‘Bell Shaped’
•Symmetrical
•Mean, Median and Mode
are Equal
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+to 
X
f(X)
μ
σ
4-6 Normal Distribution

12. 4-6 Normal Distribution
Definition

13. 4-6 Normal Distribution
Figure 4-10Normal probability density functions for
selected values of the parameters and s
2
.

14. f(X)
X
μμ+1σμ-1σ
σσ
68.26%
xμ
2σ 2σ
xμ
3σ 3σ
95.44% 99.73%

15. 4-6 Normal Distribution
Some useful results concerning the normal distribution

16. 4-6 Normal Distribution
Definition : Standard Normal
z= (x-)/s

17. 4-6 Normal Distribution
Example 4-11
Figure 4-13Standard normal probability density function.
Normal distribution table is
in Page 718 and Page 719
P(Z<=1.5)=0.93319 P(Z<=1.53)=0.93699

18. P(Z<=0.54 =0.7054)
P(Z<=1.0 =0.8413)

19. P(Z<=-2.16 =0.015)
P(Z<=-0.91 =0.1814)

20. Text-book problem
4.49
a. P(Z<1.32) =0.9066
c. P(Z>1.45) = 1-P(Z<1.45) (Because the
table gives the lower value of Z i.e., P(Z<z)
= 0.0735
d. P(Z>-2.15) = 1-0.0158=0.9842
e. P(-2.34<Z<1.76) ----see in the next slide
1.45
1 = 100%
Non-shaded area
= 1-shaded area

21. f(X)
X
1.76
-2.34
a= P(Z<1.76)
P(-2.34<Z<1.76)
b= P(Z<-2.34)
P(-2.34<Z<1.76) = P(Z<1.76)-P(Z<-2.34)
= 0.9511

22. Thanks a lot for listening.
Remaining portion we will
discuss in the next class