Uniform distribution, Slides of Quantitative Techniques

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Continuous Probability

Distributions

BS Agri 2

Group 5

Uniform Distribution

 A random variable X is said to be uniformly distribution if its density function is: 𝐹 𝑥 = 1 𝑏−𝑎 𝑎 < 𝑋 < 𝑏  The distribution derives its name from the fact that density is constant over is the or uniform interval [a,b] so why this distribution known that as rectangular & distribution.  It has 2 parameters a and b.

(i). 𝑃 𝑥 > 3 =? 𝑃 𝑥 > 3 = 3 ∞ ∫ 1 𝛽

ൗ −𝑋 𝛽 ⅆ𝑥, 𝑥 > 3 𝑃 𝑥 > 3 = 3 ∞ ∫ 1 3

−𝑋Τ (^3) ⅆ𝑥 𝑃 𝑥 > 3 = 3 ∞ ∫ 1 3

−𝑋Τ (^3) (− 3 ) ∵ 𝑒−𝜆𝑥^ ⟹ _ 1 3 𝑃 𝑥 > 3 = 3 ∞ ∫ −𝑒 Τ −𝑋 3 𝑃 𝑥 > 3 = 0 − −𝑒 − (^3) Τ (^3) ∵ − −𝑒−^1 𝑃 𝑥 > 3 = −𝑒 − 1 𝑃 𝑥 > 3 = 0. 3678

(ii) 𝑃 𝑥 > 5 =? 𝑃 𝑥 > 5 = 5 ∞ ∫ 1 𝛽

ൗ −𝑋 𝛽 ⅆ𝑥 𝑃 𝑥 > 5 = 5 ∞ ∫ 1 3

−𝑋Τ (^3) ⅆ𝑥 𝑃 𝑥 > 5 = 5 ∞ ∫ 1 3

−𝑋Τ (^3) (− 3 ) 𝑃 𝑥 > 5 = 5 ∞ ∫ 𝑒 Τ −𝑋 3 𝑃 𝑥 > 5 = 0 − [−𝑒 Τ − 5 (^3) ] 𝑃 𝑥 > 5 = 𝑒 Τ − 5 3 𝑃 𝑥 > 5 = 𝑒 − 1. 67 𝑃 𝑥 > 5 = 0. 18824

4 = 48 𝜎 4 = 48 3 𝜎 4 = 16 Radical 4 taking on b/s 4 𝜎 4 = 4 2 4 𝝈 = 𝟐

Example 9.3 If the p.d.f. of the r.v. X is 𝑓 𝑥 = 1 32 𝜋 𝑒 ൗ −(𝑥+ 7 ) 2 (^32) −∞ < 𝑥 < ∞ find its mean, variance and moment generating function. Solution: 𝑓 𝑥 = 1 32 𝜋

ൗ −(𝑥+ 7 ) 2 32 𝑓 𝑥 = 1 𝜎 2 𝜋

− 1 2 ( 𝑥−𝜇 𝜎 ) 2 ∵ 32 = 2 ( 4 ) 2

Example 9.4 Let X = N(0, 1) mean that X has a normal distribution with zero mean and unit variance. What will be the distribution of 2X - 3; 7 8 X + 5 and 4X? Solution : 𝑁 0 , 1 = 𝑁(𝜇, 𝜎 2 ) ∵ 𝜇 = 𝐸 𝑥 = 0 , 𝜎 2 = 𝑣𝑎𝑟 𝑥 = 1 ▪ Let 2x-3=y Mean: 𝐸 𝑦 = 2𝐸 𝑥 − 3

2 𝑣𝑎𝑟 𝑥) − 𝑣𝑎𝑟( 3 ∵ 𝑣𝑎𝑟 𝑥 = 1 𝑉𝑎𝑟 𝑦 = 4 1 − 0 ∵ 𝑣𝑎𝑟 𝐶𝑜𝑛𝑠𝑡 = 0 𝑽𝒂𝒓 𝒚 = 𝟒 According to N.D 𝑵(−𝟑, 𝟒).

▪ Variance: v𝑎𝑟 𝑦 = 𝑣𝑎𝑟[ 7 8

𝑥 + 5 ]

v𝑎𝑟 𝑦 = 49 64

49 64

49 64 According to N.D 𝑵 𝝁, 𝝈 𝟐 = 𝑵(𝟓, 𝟒𝟗 𝟔𝟒

 Let 4𝑥 = 𝑦 Mean: 𝐸 𝑦 = 4𝐸 𝑥 𝐸 𝑦 = 4 0 ⟹ 𝐸 𝑦 = 4 Variance: v𝑎𝑟 𝑦 = 𝑣𝑎𝑟 4 𝑥 v𝑎𝑟 𝑦 = 16 𝑣𝑎𝑟 𝑥 ∵ 𝑣𝑎𝑟 𝑥 = 1 v𝑎𝑟 𝑦 = 16 According to N.D 𝑵 𝝁, 𝝈 𝟐 = 𝑵(𝟎, 𝟏𝟔).