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The daily water consumption for an Ohio community is normally distributed with a mean consumption of 722,500 gallons and a standard deviation of 57,429 gallons. The community water system will experience a noticeable drop in water pressure when the daily water consumption exceeds 866,475 gallons. What is the probability of experiencing such a drop in water pressure? (Round your answer to 4 decimal places. Round Z-scores to 2 decimal places.)

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rodolforodriguezr

Answer:

0.0060

Step-by-step explanation:

The z-score is given by

z = (866,475-722,500)/57,429 = 2.51

The probability is given by the area under the Normal curve N(0,1) to the right of 2.51

In Excel this value is found with the formula

=1-NORMDIST(2.51,0,1,1)

and in OpenOffice Calc

=1-NORMDIST(2.51;0;1;1)

(NORMDIST(2.51;0;1;1) gives the area to the left of 2.51, so 1-NORMDIST(2.51;0;1;1) gives the area to the right of 2.51)

and equals 0.0060

You can convert the normal distribution to standard normal distribution and then use the z scores and z tables to find the needed probability.

The probability of experiencing such a drop in water pressure is 0.006 approx.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution(normal distribution whose mean is 0 and standard deviation is 1) and its values will give us the z score.

If we have [tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

Having z score will lead us to the p values (probabilities from the left end of the distribution to that z score on x axis) from the z tables.

Lets suppose the daily water consumption is measured by random variable X.

Then according to the given data, we have [tex]X \sim N(722500, 57429)[/tex]

The probability of experiencing a drop in water pressure is given by

[tex]P(X > 866475)[/tex]

(since drop in water pressure will occur if daily water consumption X exceeds 866475 gallons)

Converting this distribution to standard normal distribution, we get:

[tex]Z = \dfrac{X - 722500}{57429}\\\\Z \sim N(0,1)[/tex]

Thus, the same probability but now with Z can be expressed as

[tex]P(X > 866475) = P(Z > \dfrac{866475-722500}{57429}) \approx P(Z > 2.507)[/tex]
or

[tex]P(Z > 2.507) = 1 - P(Z \leq 2.507)[/tex]

Referring to z tables to get p value for z = 2.507, we get 0.994

p value shows [tex]P(Z \leq z) = (\text{p value for Z = z})[/tex]

Thus, we have [tex]P(Z \leq 2.507) = 0.994[/tex]
Thus,

[tex]P(Z > 2.507) = 1 - P(Z \leq 2.507) = 1 - 0.994 = 0.006\\P(X > 866475) = P(Z > 2,507) = 0.006[/tex]

Thus,

The probability of experiencing such a drop in water pressure is 0.006 approx.

Learn more about z score here:

https://brainly.com/question/21262765

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