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
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 ![\mu = 10, \sigma = 1.5](https://tex.z-dn.net/?f=%5Cmu%20%3D%2010%2C%20%5Csigma%20%3D%201.5)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{11.5 - 10}{1.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B11.5%20-%2010%7D%7B1.5%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
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.