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Standard Normal Distribution • Normal Distribution f ( x) 1 2 e ( x )2 2 , X When x = 0 and σ = 1 then f ( x) 1 2 e ( x)2 2 , X 2 A normal random variable with 0 and 1 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 P( x1 X x 2 ) x2 x1 1 2 e 1 ( x )2 2 dx 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: P( z1 Z z 2 ) z2 z1 Z 1 2 X e 1 ( x )2 2 dz 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 1. 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 = 0.52k 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 k 2.37 Standard Normal Distribution • Example 4: • Given a random variable X having a normal distribution with 50 and 10 , 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 X 45 50 z1 0.5 10 X 62 50 z2 1 .2 10 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) z X 362 300 1.24 50 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: 1. 45% of the area to left. 2. 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 z X X z 6 (0.13) 40 39.22 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 X z X z 6 (1.08) 40 46.48