Question

A random sample of 1,700 cars is checked for a serious defect. The chance that a car has the defect is 0.6%. Approximate the probability that at least 10 cars in the sample have the defect. State and check the appropriate conditions.

Answer 1

52.39% probability that at least 10 cars in the sample have the defect.

Conditions:

Each car has only two possible outcomes. Either it is defective, or it is not. The cars are chosen at random, which means that the probability of a car being defective is independent from other cars. So we use the binomial probability distribution to solve this questions.

Explanation:

Conditions:

Each car has only two possible outcomes. Either it is defective, or it is not. The cars are chosen at random, which means that the probability of a car being defective is independent from other cars. So we use the binomial probability distribution to solve this questions.

We are working with a large sample, which means that we can approximate the binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`n = 1700, p = 0.006`

So

`E(X) = np = 1700*0.006 = 10.2`

`\sigma = √(V(X)) = √(np(1-p)) = √(1700*0.006*0.994) = 3.18`

Approximate the probability that at least 10 cars in the sample have the defect.

This is 1 subtracted by the pvalue of Z when X = 10. So

`Z = (X - \mu)/(\sigma)`

`Z = (9 - 10.2)/(3.18)`

`Z = -0.06`

`Z = -0.06` has a pvalue of 0.4761.

1 - 0.4761 = 0.5239

52.39% probability that at least 10 cars in the sample have the defect.