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

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 and standard deviation , the z-score of a measure X is given by:

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

Proportion of days with the daily rainfall above 11.5 mm.

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

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.

step-by-step explanation: