Question 130976: I need help with this question. Can you please help me solve this question step-by-step.
Question: A resistor in a certain circuit is specified to have a resistance in the range 99 Ω – 101 Ω. An engineer obtains two resistors. The probability that both of them meet the specification is 0.36, the probability that exactly one of them meets the specification is 0.48, and the probability that neither of them meets the specification is 0.16. Let X represents the number of resistors that meet the specification. Find 1. the probability distribution function 2). the mean and variance of X.
Thanks.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Question: A resistor in a certain circuit is specified to have a resistance in the range 99 Ω – 101 Ω. An engineer obtains two resistors.
The probability that both of them meet the specification is 0.36
P(2 successes) = p^2 = 0.36 implies p = 0.6 and q = 0.4
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the probability that exactly one of them meets the specification is 0.48
P(exactly one success) = 2C1pq = 2pq = 0.48 implies pq = 0.24; p=0.6,q=0.4
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and the probability that neither of them meets the specification is 0.16.
P(no successes) = q^2 = 0.16 implies q = 0.4
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Let X represents the number of resistors that meet the specification.
Find 1. the probability distribution function
X is the random variable and is a count of successes in two tries:
X values are 0, 1, 2
P(0) = q^2 = 0.16
P(1) = 2pq = 0.48
P(2) = p^2 = 0.36
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2). the mean and variance of X.
mean = np = 2*0.6 = 1.2
variance = npq = 1.2*0.4 = 0.48
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Cheers,
Stan H.
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