Question

The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of 4.5. a) Find the probability that in a given year there will be less than 21 earthquakes. b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

Answer 1

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Explanation:

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.

In this problem, we have that:

`\mu = 20.8, \sigma = 4.5`

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

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

`Z = (21 - 20.8)/(4.5)`

`Z = 0.04`

`Z = 0.04` has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

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

`Z = (23 - 20.8)/(4.5)`

`Z = 0.71`

`Z = 0.71` has a pvalue of 0.7611

X = 18

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

`Z = (18 - 20.8)/(4.5)`

`Z = -0.62`

`Z = -0.62` has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.