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.

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:

Standard Normal Distribution Areas under the normal curve:-

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

Standard Normal Distribution as:

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

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 = -1.97 and z = 0.86.

Standard Normal Distribution Example 2:

Standard Normal Distribution Area left z = 1.84 = 1 – area right of z = 1.84 = 1 – 0.9671 = 0.0329 2. Area between z = -1.97 and z = 0.86 is equal to the area of z = 0.86 – area of z = - 1.96 = 0.8051 – 0.0244 = 0.7807

Standard Normal Distribution Example 3: Given a standard normal distribution, find the value of k such that: P (z > k) = 0.3015 P (k < z < - 0.18) = 0.4197

Standard Normal Distribution Example 3:

Standard Normal Distribution P (z > k) = 1 - P (z < k) = 1 – 0.3015 = 0.6985 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 = 0.4286 نبحث عن هذه المساحة في الجدول ونجد أن z = - 2.37

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.

Standard Normal Distribution Example 4:

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) = 0.8849 – 0.3085 = 0.5764

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

Standard Normal Distribution Solution: P (X > 362) = 1 – P (X < 362) Then P(X>362)= 1 – P (z < 1.24) = 1 – 0.8925= 0.1075

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.

Standard Normal Distribution Example 6:

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

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