The probability of a car being defective is 1/10. In a survey of 150 cars, what is probability that 20 or more cars are defective?

Two ways to tackle this. The hard way is to determine the probability of 20 being defective, plus the probabiity 21 being defectice, plus the probability of 22 being defecetive , plus... etc. Each of these probabilities is calculated using the formula for binomial distribution:

P(n) = C(150,n)p^n q^{(150-n)}

An easier way is to use the fact that the binomial distribution approaches a normal distribution for large populations. So you can use the cumulative probability for a normal distribution, where \mu = mean = np, \sigma = standard deviation = \sqrt{npq}, and you want to check a z-score of (x-\mu )/ \sigma for x = 20.

Many thanks for your help