Suppose that only 0.1% of all computers of a certain type experience CPU failure

Question

Suppose that only 0.1% of all computers of a certain type experience CPU failure during the warranty period. Consider a sample of 7,000 computers.

(a) What are the expected value and standard deviation of the number of computers in the sample that have the defect? (Round your standard deviation to two decimal places.) expected value computers standard deviation computers

(b) What is the (approximate) probability that more than 9 sampled computers have the defect? (Round your answer to three decimal places.)

(c) What is the (approximate) probability that no sampled computers have the defect? (Round your answer to five decimal places.)

Explanation / Answer

the probability of experiencing failure = 0.1

probability of no failure = 0.9

now we total have 7000 computers

therefore expectation =

E(X) = X*P(X) = 7000*0.1 = 700

VAR(X) = E(X^2) - E(X)^2

= 0.1*7000^2 - 700^2

= 4410000

NOW THE STANDARD DEVIATION = 4410000^(1/2) = 2100

B) HERE WE NEED TO FIND THAT AMONG 7000

MORE THAN 9 ARE DEFECTIVE

P(X>9) = 1 -[P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)]

P(0) = 7000C0*(0.1)^0*(0.9)^7000 = 0.00

SIMILARLY WE WILL CALCULATE TILL P(9)

THA ANSWER WILL COME OUT AFTER DOING OUT CALCULATION ON COMPUTER = 1 - 0.12 = 0.88

C) P(X=0) = 7000C0*(0.1)^0*(0.9)^7000 = 0.00

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