Respuesta :
Using the normal distribution, it is found that:
a) There is a 0.1867 = 18.67% probability that a day requires more water than is stored in city reservoirs.
b) A capacity of 415 million gallons is needed.
c) An amount of 236 million gallons of water use is exceeded with 95% probability.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation for this problem are given as follows:
[tex]\mu = 310, \sigma = 45[/tex]
For item a, the probability is one subtracted by the p-value of Z when X = 350, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (350 - 310)/45
Z = 0.89
Z = 0.89 has a p-value of 0.8133.
1 - 0.8133 = 0.1867.
There is a 0.1867 = 18.67% probability that a day requires more water than is stored in city reservoirs.
For item b, the capacity should be the 99th percentile, which is X when Z = 2.327, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
2.327 = (X - 310)/45
X - 310 = 45 x 2.327
X = 415. (rounded)
A capacity of 415 million gallons is needed.
For item c, the amount of water use is the 5th percentile, which is X when Z = -1.645, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
-1.645 = (X - 310)/45
X - 310 = -1.645 x 45
X = 236.
An amount of 236 million gallons of water use is exceeded with 95% probability.
More can be learned about the normal distribution at https://brainly.com/question/15181104
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