1. Standard Normal Distribution
When x = 0 and σ = 1 then
A normal random variable with
is called a standard normal variable and it’s denoted as Z.

2. Standard Normal Distribution
Areas under the normal curve:-
The curve of any continuous probability function or density function is constructed by integrate the function between x1, x2 as:

3. Standard Normal Distribution
Areas under the normal curve:-

4. Standard Normal Distribution
The above function can be transforming
all the observations of it into a new set of
observations to a normal random variable
Z with mean = 0 and variance = 1, and this
transformation written

5. Standard Normal Distribution
as:

6. Standard Normal Distribution
Find
P( z < 1.74) = 09591

7. Standard Normal Distribution
Example 2:
Given a standard normal distribution, find the area under the curve lies:
To the right of z=1.84.
Between z = and z = 0.86.

8. Standard Normal Distribution
Example 2:

9. Standard Normal Distribution
Area left z = 1.84 = 1 – area right of
z = 1.84 = 1 – =
2. Area between z = and z = 0.86 is equal to the area of z = 0.86 – area of
z = = – =

10. Standard Normal Distribution
Example 3:
Given a standard normal distribution, find the value of k such that:
P (z > k) =
P (k < z < ) =

11. Standard Normal Distribution
Example 3:

12. Standard Normal Distribution
P (z > k) = 1 - P (z < k) = 1 – =
نبحث في الجدول عن هذه المساحة ونجد أنها تقابل القيمة
Z = 0.52
P (k < z < -0.18) we should find the area of z = 0.18 =
نبحث عن هذه المساحة في الجدول ونجد أن
z =

13. Standard Normal Distribution
Example 4:
Given a random variable X having a normal distribution with
, find the probability that X assumes a value between 45 and 62.

14. Standard Normal Distribution
Example 4:

15. Standard Normal Distribution
Solution
x1 = 45 , x2 = 62
P (45 < X < 62) = P ( -0.5 < z < 1.2)
Then
P ( -0.5 < z < 1.2) = P (z<1.2) – P(z < -0.5)
= – =

16. Standard Normal Distribution
Example 5:
Given that X has a normal distribution with find the probability that X assumes a value greater than 362.

17. Standard Normal Distribution
Solution:
P (X > 362) = 1 – P (X < 362)
Then
P(X>362)= 1 – P (z < 1.24)
= 1 – =

18. Standard Normal Distribution
Example 6:
Given that a normal with find the value of (x) that has:
45% of the area to left.
14% of the area to the right.

19. Standard Normal Distribution
Example 6:

20. Standard Normal Distribution
Solution:
P (z < k) = 0.45
Then k = -0.13

21. Standard Normal Distribution
Solution:
14% of the area to the right = 1 – the area of 14% of the left = 1 – 0.14 = 0.85
P (z < k) = 0.86
Then k = 1.08