A contractor purchases a shipment of 99 transistors. The way the business works is that the contractor will randomly select and test 10 of these transistors, and keep the shipment only if at least 9 of the 10 are in working condition. If we know that overall, 20% of the transistors have defects (and thus not in working condition), then what is the probability the contractor will keep all the transistors?

the probability that the contractor keeps the transistors is 37.58%

Step-by-step explanation:

Since each transistor is equally likely to the chosen , then the random variable X= transistors that have defects from a sample of 10 , has a binomial distribution . Thus its probability is calculated through:

P(X=x) = n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= sample size = 10

p= probability for any transistor to be defective = 0.2 (20%)

x= number of transistors that are defective

P(X=x) = probability that there are x transistors that are defective

thus replacing values and knowing that only 0 or 1 can be defective in order to have at least 9 working transistors:

P(X≤1)= P(X=1)+P(X=0)

this is equivalent to

P(X≤1)= F(X=1)

where F(X) is the cummulative binomial probability distribution. From tables can be found that

P(X≤1)= F(X=1)= 0.3758 (37.58%)

thus the probability that the contractor keeps the transistors is 37.58%

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation:

first, we define the variables:

x: number of years after 1950

f (x): amount of vinyl sold.

then, with the variables defined, we have:

68594 vinyl records were sold in 1958 > f (8) = 68594

91299 vinyl records were sold in 1961 > f (11) = 91299

38720 vinyl records were sold in 1952 > f (2) = 38720

161743 vinyl records were sold in 1967 > f (17) = 161743

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