An electronics store has received a shipment of 20 table radios that have connec

Question

An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the other eight have a single slot. Suppose that five of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let X = the number among the radios stored under the display shelf that have two slots.

(b) Compute

P(X = 2), P(X 2), and P(X 2).

(Round your answers to four decimal places.)

(c) Calculate the mean value and standard deviation of X. (Round your standard deviation to two decimal places.)

Explanation / Answer

PROBABILITY OF TWO SLOT = 12/20 = 0.6

PROBABILITY OF ONE SLOT = 0.4

X = TWO SLOT RADIO

N = 5

B) P( X=2) = 5C2*(0.6)^2*(0.4)^3 = 0.230

P(X<=2) = 1 - P(0)-P(1)

P(0) = 5C0*(0.6)^0*(0.4)^5 = 0.010

P(1) = 5C1*(0.6)^1*(0.4)^4 = 0.076

THEREFORE P(X<=2) = 1 - 0.086 = 0.914

P(X>=2) = 1 - P(X<=2) = 1 - 0.914 + P(2) = 1 - 0.914+0.230 = 0.316

B) E(X) = 0*P(0)+1*P(1)+2*P(2)+3*p(3)+4*P(4)+5*P(5)

P(3) = 5C3*(0.6)^3*(0.4)^2 = 0.345

P(4) = 5C4*(0.5)^4*(0.4)^1 = 0.064

P(5) = 0.6^5 =0.077

THEREFORE MEAN = 0*0.010+1*0.076+2*0.230+3*0.345+4*0.064+5*0.077 = 2.12

VAR(X) = E(X^2) - E(X)^2 = 7.05 - 4.49 = 2.56

HENCE STANDARD DEVIATION = VAR(X)^(1/2) = 2.56^(1/2) = 1.6

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