1. Four percent of DVD players manufactured by an electronics company are defect

Question

1. Four percent of DVD players manufactured by an electronics company are defective. Suppose six DVD players are randomly selected from the production line. Each DVD player is determined to be defective or non-defective. a. What is the probability that none of the DVD players are defective? b. What is the probability that all six of the DVD players are defective? c. What is the probability that at most one of the DVD players is defective? d. What is the probability that at least two of the ovD players are defective? T Ctri)-

Explanation / Answer

1.

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 6 * 0.04
= 0.24
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 6 * 0.04 * 0.96
= 0.2304
III.
standard deviation = sqrt( variance ) = sqrt(0.2304)
=0.48
a.
probability that none of the DVD players defective
P( X = 0 ) = ( 6 0 ) * ( 0.04^0) * ( 1 - 0.04 )^6
= 0.7828
b.
probability that all of six DVD players are defective
P( X = 6 ) = ( 6 6 ) * ( 0.04^6) * ( 1 - 0.04 )^0
= 0
c.
probability that atmost 1 of the DVD player are defective
P( X < = 1) = P(X=1) + P(X=0) +
= ( 6 1 ) * 0.04^1 * ( 1- 0.04 ) ^5 + ( 6 0 ) * 0.04^0 * ( 1- 0.04 ) ^6 +   
= 0.9784
d.
probability that atleast 2 of DVD player are defective
P( X < 2) = P(X=1) + P(X=0) +   
= ( 6 1 ) * 0.04^1 * ( 1- 0.04 ) ^5 + ( 6 0 ) * 0.04^0 * ( 1- 0.04 ) ^6 +
= 0.9784
P( X > = 2 ) = 1 - P( X < 2) = 0.0216

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