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# Continuous Random variable - Class 3.pdf

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### Continuous Random variable - Class 3.pdf

1. 1. Continuous Random variable Introduction Continuous random variables can represent any value within a specified range or interval and can take on an infinite number of possible values. If the image is uncountably infinite (usually an interval) then X is called a continuous random variable. In the special case that it is absolutely continuous, its distribution can be described by a probability density function(pdf), which assigns probabilities to intervals; in particular, each individual point must necessarily have probability zero for an absolutely continuous random variable An example of a continuous random variable would be an experiment that involves measuring the amount of rainfall in a city over a year or the average height of a random group of 25 people. Probability density function: A function f, defined for all x in (−∞ ∞,) is called the probability density function of a continuous random variable X if i) f(x) ≥0, for all x in (−∞ ∞,) ∞ ∫= 1 ii) f x dx ( ) −∞ Note: a ∫ 1. P(X < a) = ( ) f x dx −∞ ∞ ∫ 2. P(X > a) = ( ) f x dx a b f x dx ∫ 3. P ( a < X < b) = ( ) a CDF
2. 2. x ∫ F(x) = P(X ≤x) = ( ) f x dx −∞ Check your progress 1 π 1+ x , −∞ ≤x≤ ∞can be the p.d.f of a continuous RV. 1. Examine f(x) = ( )2 ∞ 1 π 1+ x dx = 1 πTan-1 (x)} ∫( )2 −∞ = 1 π( 2 π - (-2 π )) = 1 2. If f(x) = k(1 + x) in 2 < x < 5 is the p.d.f of a continuous RV X, Find i) P(X < 4) ii) P(3 < X < 4) iii) P(3 < X < 6) Solution: Since f(x) is the pdf ∞ ∫= 1 f x dx ( ) −∞ ∞ ∫ 1 = f x dx ( ) −∞ = 2 ∫+ f x dx ( ) −∞ 5 5 f x dx ( ) ∫+ 2 ∞ ∫ 5 f x dx ( ) = 0 + k (1 ) + x dx ∫+ 0 2 = k (1 + x)2 /2 }UL = 5, LL = 2 = k/2{(1+5)2 – (1 + 2)2 } = k/2{ 36 – 9} 1 = (k/2)(27) ⇨ K = 2/27 4 ∫
3. 3. i) P(X < 4) = = f x dx ( ) −∞ 2 ∫+ f x dx ( ) −∞ 4 4 f x dx ( ) ∫ 2 = 0 + 2/27 (1 ) + x dx ∫ 2 = 2/27 { (1+x)2 /2} UL = 4, LL =2 =1/27{(1+4)2 - (1+2)2 } = 1/27 { 25 -9} =16/27 4 ii) P(3 < X < 4)= 2/27 (1 ) + x dx ∫ 3 = 2/27 {(1+x)2 /2} UL = 4, LL = 3 = 1/27 {25 – 16} = 9/27 = 1/3 1 2 3 4 5 2/27(1+x) iii) P(3 < X < 6) = P(3 < X < 5) + P(5<X <6) 5 = 2/27 (1 ) + x dx ∫ 3 = 2/27 {(1+x)2 /2}UL = 5, LL= 3 = 1/27{ 36 – 16} = 20/27 3. The amount of bread (in hundreds of kgs) that a certain bakery is able to sell in a day is a random variable X with a pdf by ⎧ ≤ < ⎪ ⎨ − ≤ < A x if x , 0 5 A x x f(x) =
4. 4. ⎪ ⎩ 0 , (10 ), 5 10 other wise i) Find the value of A ii) Find the probability that in a day the sales is (a) More than 500 kg (b) less than 500 kgs (c) between 250 and 750 kgs iii) Find P(X >5/X ≤5) iv) P(X > 5/ 2.5 <X < 7.5) Solution: Since f(x) is a pdf ∞ ∫ 1 = f x dx ( ) −∞ 5 = 0 ∫+ f x dx ( ) f x dx ( ) ∫+10 ∞ ∫ f x dx ( ) ∫+ f x dx ( ) −∞ 5 0 10 5 10 = 0 +A x dx ∫+ A 0 (10 ) − x dx ∫+0 5 = A{ x2 /2] UL =5, LL=0 + (10-x)2 /2(-1)] UL =10, LL=5} = A{ ½(25 – 0) -1/2(0 – 25)} = A{ 25} ⇨ A = 1/25 ∞ ∫= P(5<X<10)+P(10<X<00) ii) a) P(X >5)= 5 10 f x dx ( ) = f x dx ( ) ∫+0 5 = 1/25 10 (10 ) − x dx ∫ 5 = 1/25 { (10 –x)2 /(-2)} UL = 10, LL=5 = -1/50{ 0 – 25} = ½
5. 5. b) P(X<5) = 1 - P(X >5) = 1 – ½ = ½ c) P(2.5<X<7.5) = P(2.5<X<5) +(5<X<7.5) = 1/25 5 ( ) x dx ∫+ 1/25 2.5 7.5 (10 ) − x dx ∫ 5 = 1/25{ x2 /2 ] UL = 5, LL= 2.5+ (10-x)2 /(-2)]UL =7.5, LL =5} = 1/25{1/2( 25 – 6.25) -1/2(6.25 – 25)} = 1/50{ 18.75 +18.75} = ¾ iii) P(X >5/X ≤5) = P(X>5 and X≤5)/P(X≤5) = 0 P(A/B) = P(A and B)/P(B) 1 2 3 4 5 6 7 8 9 10 iv) P(X > 5/ 2.5 <X < 7.5) = P(X>5 and 2.5 <X < 7.5) / P(2.5 <X < 7.5) 0 1 2 3 4 5 6 7 8 9 10 = P(5<X<7.5)/ P(2.5 <X < 7.5) = (3/8) / (3/4) = 1/2