Question

ion of a random variable, X, is a nonnegative Definition 1. The probability denisty function of a random variable function fx with the property that (a) for all values x, P(X = x) = fx(x) if X is discrete and (b) for all intervals [a, b], P(a < X <b) is the area under the curve of b if X is continuous. Recall the experiement from Example 2.6-6 from last class: At 25°C, 20% of a certain type of laser diodes have efficiency below 0.3 mw/mA. Five diodes are selected by simple random sampling from a large population of such diodes. 1. Find the probability that exactly three diodes will have efficiency below 0.3 at 25°C. 2. Build a random variable X for your experiement. Is X continuous or discrete? Why? 3. For each value X can take on, describe the event associated with that value. 4. Build the probability denisty function fx.

Answer 1

P( Efficiency below 0.3) = 0.2 =p

n=5

X be the number of diode that have efficiency below 0.3

X~ Binomial ( n=5, p=0.2)

a) P( X=3) = ⁵C₃ (0.2)³ (1-0.2)² = 0.0512

b) X is discrete , as the number of diodes cannot be in fraction it will be either0 or 1 or 2 or 3 or 4 or 5,.

X = { 0,1,2,3,4,5}

c) For X =0, it means that out of 5 no diode have efficiency below 0.3

   For X =1, it means that out of 5 only 1 diode have efficiency below 0.3

For X =2, it means that out of 5 only 2 diode have efficiency below 0.3

For X =3, it means that out of 5 only 3 diode have efficiency below 0.3

For X =4, it means that out of 5 all 4 diode have efficiency below 0.3

  For X =5, it means that all 5 diode have efficiency below 0.3

d) Probability density function

X	0	1	2	3	4	5
P(X=x)	5C0 (0.2)0 (1-0.2)5 =0.32768	5C1 (0.2)1 (1-0.2)4 = 0.4096	5C2 (0.2)2 (1-0.2)3 = 0.2048	5C3 (0.2)3 (1-0.2)2 = 0.0512	5C4 (0.2)4 (1-0.2)1 =0.0064	5C5 (0.2)5 (1-0.2)0 =0.00032