Question

About 12.5% of restaurant bills are incorrect. If 200 bills are selected at ran- dom, find the probability that at least 22 will contain an error. Is this likely or unlikely to occur

Answer 1

0.7734 = 77.34% probability that at least 22 will contain an error. Probability above 50%, which means that this is likely to occur.

Explanation:

Binomial probability distribution

Probability of exactly x successes 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 distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, 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))`.

About 12.5% of restaurant bills are incorrect.

This means that
`p = 0.125`

200 bills are selected at random

This means that
`n = 200`

Mean and standard deviation:

`\mu = E(X) = np = 200*0.125 = 25`

`\sigma = √(V(X)) = √(np(1-p)) = √(200*0.125*0.875) = 4.677`

Find the probability that at least 22 will contain an error.

Using continuity correction, this is
`P(X \geq 22 - 0.5) = P(X \geq 21.5)`, which is 1 subtracted by the p-value of Z when X = 21.5. So

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

`Z = (21.5 - 25)/(4.677)`

`Z = -0.75`

`Z = -0.75` has a p-value of 0.2266.

1 - 0.2266 = 0.7734

0.7734 = 77.34% probability that at least 22 will contain an error. Probability above 50%, which means that this is likely to occur.