Question

In Waterville, the average daily rainfall in July is 10 mm with a standard deviation of 1.5 mm. Assume that this data is normally distributed. How many days in July would you expect the daily rainfall to be more than 11.5 mm

Answer 1

Answer:

You should expect 5 days in July with daily rainfall of more than 11.5 mm.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal 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:

$Z = \frac{X - \mu}{\sigma}$

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.

In Waterville, the average daily rainfall in July is 10 mm with a standard deviation of 1.5 mm.

This means that $\mu = 10, \sigma = 1.5$

Proportion of days with the daily rainfall above 11.5 mm.

1 subtracted by the p-value of Z when X = 11.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{11.5 - 10}{1.5}$

$Z = 1$

$Z = 1$ has a p-value of 0.84.

1 - 0.84 = 0.16.

How many days in July would you expect the daily rainfall to be more than 11.5 mm?

July has 31 days, so this is 0.16 of 31.

0.16*31 = 4.96, rounding to the nearest whole number, 5.

You should expect 5 days in July with daily rainfall of more than 11.5 mm.