Question

Suppose that the number of bikes rented in a week from the bike shop has mean 500and standard deviation 20. Assume that the number of bikes rented in different weeks isindependent. Find the probability that the average number of bikes rented per week, duringthe next 30 weeks, will be at least 510.

Answer 1

0.31% probability that the average number of bikes rented per week, during the next 30 weeks, will be at least 510.

Explanation:

To solve this question, we need 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 random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 500, \sigma = 20, n = 30, s = (20)/(√(30)) = 3.65`

Find the probability that the average number of bikes rented per week, during the next 30 weeks, will be at least 510.

This is 1 subtracted by the pvalue of Z when X = 510. So

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

By the Central Limit Theorem

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

`Z = (510 - 500)/(3.65)`

`Z = 2.74`

`Z = 2.74` has a pvalue of 0.9969

1 - 0.9969 = 0.0031

0.31% probability that the average number of bikes rented per week, during the next 30 weeks, will be at least 510.