Question

A recent survey found that about 72.5% of all gasoline purchases at a certain service station chain are paid with a credit or debit card. If 300 gasoline purchases completed at this chain are randomly selected, find the probability that at most 210 of those purchases are paid with a credit or debit card.

Answer 1

18.28% probability that at most 210 of those purchases are paid with a credit or debit card.

Explanation:

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 = 300, p = 0.725`

So

`\mu = E(X) = np = 300*0.725 = 217.5`

`\sigma = √(V(X)) = √(np(1-p)) = √(300*0.725*0.275) = 7.7339`

Find the probability that at most 210 of those purchases are paid with a credit or debit card.

Using continuity correction, this is
`P(X \leq 210 + 0.5) = P(X \leq 210.5)`, which is 1 subtracted by the pvalue of Z when X = 210.5. So

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

`Z = (210.5 - 217.5)/(7.7339)`

`Z = -0.905`

`Z = -0.905` has a pvalue of 0.1828

18.28% probability that at most 210 of those purchases are paid with a credit or debit card.