Ms.
Lubna Ineirat
108
Test result
Yes (D)
No(
ഥ
𝐷
)
Total
Positive
436
5
Negative
14
495
Total

3. What is the false positive.
4. What is the false negative .
5.
suppose its known that the rate of disease its 11.3% find the
predictive value positive.
6. suppose its known that the rate of disease its 11.3% find the
predictive value negative.
Ms.
Lubna Ineirat
109

Chapter
5:
Standard
Normal distribution
Properties of standard
normal (Z)
1.
The curve is of bell shape and symmetric about 0.
2.
The area under the curve is 1.
3.
Mean=median=mode=0
4.
Variance=1
Ms.
Lubna Ineirat
110

Ms.
Lubna Ineirat
111
Using Z -table to find the area

Using Z -table to find the area
Case 1:
less than
•
P(Z<1.59)
•
P(Z<2.31)
•
P(Z<2.08)
•
P(Z<0.07)
•
P(Z<3)
•
P(Z<0)
•
P(Z<3.86)
•
P(Z<5)
•
P(Z<8)
Ms.
Lubna Ineirat
112

Ms.
Lubna Ineirat
113
Find the following:
•
P(Z<-1.39)
•
P(Z<-2.31)
•
P(Z<-0.05)
•
P(Z<-2.06)
•
P(Z<-1.5)
•
P(Z<-0.8)
•
P(Z<-0.24)
•
P(Z<-4)
•
P(Z<-3.45)
•
P(Z<-6)
•
P(Z<-11)

Case 2:
more than
Ms.
Lubna Ineirat
114
•
P(Z>1.59)
•
P(Z>2.31)
•
P(Z>1.08)
•
P(Z>0.07)
•
P(Z>2.3)
•
P(Z>0.57)
•
P(Z>3)
•
P(Z>0)
•
P(Z>3.86)
•
P(Z>5)
•
P(Z>8)

Ms.
Lubna Ineirat
115
Find the following
•
P(Z>-3.59)
•
P(Z>-1.51)
•
P(Z>-2.02)
•
P(Z>-0.04)
•
P(Z>-6.2)
•
P(Z>-0.87)
•
P(Z>-3)
•
P(Z>-3.86)
•
P(Z>-5)
•
P(Z>-7)

Case 3:
more than
& less than (between)
P(1.54< Z< 2.39)
---------------------------
P(0< Z <3.49)
-------------------------
P(2.56 < Z< 3.12)
--------------------------
P(-1.54< Z < 2.39)
Ms.
Lubna Ineirat
116

Find the value of c
if
Case 1:
less than
P(Z<c)=0.8413
P(Z<c)=0.9793
P(Z<c)=0.9934
P(Z<c)=0.8213)
Ms.
Lubna Ineirat
117

Find the value of c
if
P(Z<c)=0.0022
P(Z<c)=0.0158
P(Z<c)=0.1711
P(Z<c)=0.1210)
P(Z<c)=0.9726
Ms.
Lubna Ineirat
118

Find the value of c
Case 2:
more than
•
P(Z>c)=0.9821
•
P(Z>c)=0.9948
•
P(Z>c)=0.0030
•
P(Z>c)=0.9251
•
P(Z>c)=0.505
Ms.
Lubna Ineirat
119

Find the value of c
Case 3: (between)
•
P(c<Z>0)=0.1700
•
P(0<Z>c)=0.4983
•
P(c<Z>3.54)=0.994530
•
P(-2.53<Z>c)=0.9934
Ms.
Lubna Ineirat
120

Not Standard Normal
Let X be a normal with mean
μ
and
variance
𝜎
2
(X:N(
μ
,
𝜎
2
))
Then Z=
𝑋−𝜇
𝜎
is standard normal, then Z=
𝑋−𝜇
𝜎
is N(0,1)
Not standard
Ms.
Lubna Ineirat
121

Example :
if mean =50, variance=100 find the following:
1.
P(X<60)
2.
P(X<30)
3.
P(X<90)
4.
P(X>300
5.
P(X>70)
6.
P(X>20)
7.
P(X>90)
8.
P(50<X<65)
9.
P(30<X<55)
10. P(X=70)
11. P(X=30)
Ms.
Lubna Ineirat
122

12. Find
𝑃
67
13. Find
𝑃
33
14. Find
𝑃
75
Ms.
Lubna Ineirat
123

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