1 answer

It is known that 80% of a defected computers can be repaired. A sample of 12...

Question:

It is known that 80% of a defected computers can be repaired. A sample of 12 computers is selected randomly, find the probability that;

(i) At least 3 computers can be repaired. CR [5] (ii) 5 or less can be repaired CR [5] (iii) None can be repaired CR [2] Based on your output for these, how will you advised after a critical analysis from your output.

(b) Let X have the density function f(x)= {█( @0.75t(1-x^2 ) , -1≤x≤1 @ @0, otherwise . )┤

Explain and Comment on the output of these probabilities below, what will be the conclusion of your output.

(i)compute for t EV [5] (ii)Find the probabilities; (α) P(-1⁄(4 )≤x≤1⁄3) EV [4] (β) P(1⁄2≤x≤2) EV [4]


Answers

It is known that 80% of a defected computers can be repaired. Kelo, X2 the number of computer repairied if this is a success,21 - N +(9 (0-8024 (0.2022+ (?) (0-80)$(0-20) (i) xt least 3 computer can be repaired ie P(X), 3) P(x.<3) 21- {P(x20) +P(X21O 39 P(x20) = 162) 1080) (0-20)/2 C None can be repaired. 9 2 4:096610-9 2) ket x has the density function- f(x)-50.75+ (1-x20.75 x 2 -/ NJ 1. => tz * 0.75 + {2-33-1 -> 0.75 t x x 4 = 1 3 -1. 4x0.75 So, the value of It-1 (üywe have to find the follow13 2 075 is / 14 -/y 1-1/4 dx- / xdu] -0.7531x130 - 13153 24 +) - 5 latton) -0.75 { ž - 6tz #to).</p><p>Z 0.75 + 2 100 4243 x22 (C + 1/2 - ! 468) dx +S+ frasex 04 (0-2) da - s o 12 sz { Sih dx- // 20.75 1 x de 2 0.75 {(1-6) – 12 ) } 20.75 (0.5-1 24). 2

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