210 A random sample of 10 resistors is to be tested From pas

2.10 A random sample of 10 resistors is to be tested. From past experience, it is known that the probability of a given resistor being defective is 0.08. Let X be the number of defective resistors. a. What kind of distribution function would be recommended for modeling the r.v. X? b. According to the distribution function in (a), what is the probability that in the sample of 10 resistors, there are more than 1 defective resistors in the sample?

Solution

a. This would be a binomial distribution. The number of trials is fixed, 10. The resistors are independent of each other. Each resistor has the same probability of being defective, 0.08. X would be a binomial random variable where n is 10 and p is 0.08.

b. The probability of more than 1 defective is equal to 1 minus the probability of 0 or 1 defective.

binomial probability formula:
P(X = x) = nCx*p^x*(1-p)^(n-x)

The probabilty that there is 1 defective:

P(X = 1) = 10C1*(.08)^1*(1-.08)^(10-1) = 0.3777

P(X = 0) = 10C0*(.08)^0*(1-.08)^(10-0) = 0.4344

Probability that x = 0 or 1 = .3777 + .4344 = 0.8121

The probability that x is greater than 1 = 1 - 0.8121 = 0.1879 (answer)