Question

We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all five baskets of fries?

Answer 1

32.64% probability that you would have enough money to pay for all five baskets of fries

Explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n can be approximated to a normal distribution with mean

In this problem, we have that:

`\mu = 6, \sigma = 2, n = 5, s = (2)/(√(5)) = 0.8944`

You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all five baskets of fries?

28/5 = 5.6

So this is the pvalue of Z when X = 5.6.

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

By the Central Limit Theorem

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

`Z = (5.6 - 6)/(0.8944)`

`Z = -0.45`

`Z = -0.45` has a pvalue of 0.3264

32.64% probability that you would have enough money to pay for all five baskets of fries