Question

Monthly water bills for a city have a mean of $108.43 and a standard deviation of $36.98. Find the probability that a randomly selected bill will have an amount greater than $173, which the city believes might indicate that someone is wasting water. Would a bill that size be considered unusual?

Answer 1

Answer:

The bill size can be considered usual.

Step-by-step explanation:

Mean (μ) = $108.43

Standard deviation (σ) = $36.98

Now, consider the distribution to be normal distribution :

$P(Z=\frac{X-\mu}{\sigma}>\frac{a-\mu}{\sigma})=P(Z>\frac{173-108.43}{36.98})\\\\\implies P(Z>1.75)$

Now, finding values of z-score from the table. We get,

P(Z > 1.75) = 0.9599

⇒ 95.99%

So, only 4.01 % of the people in the city wastes water.

Hence, the bill size can be considered usual.

Answer 2

Answer:

The probability that a randomly selected bill will have an amount greater than $173 is 0.04006.    

No, a bill with $173 can not be considered unusual.

Step-by-step explanation:

We have been given that monthly water bills for a city have a mean of $108.43 and a standard deviation of $36.98.

First of all, we will find z-score of data point 173 using z-score formula.

$z=\frac{x-\mu}{\sigma}$, where,

$z=\text{Z-score}$

$x=\text{Random sample score}$

$\mu=\text{Mean}$

$\sigma=\text{Standard deviation}$

Substitute the given values:

$z=\frac{173-108.43}{36.98}$

$z=\frac{64.57}{36.98}$

$z=1.75$

Now, we will use formula $P(z>a)=1-P(z$ as:

$P(z>1.75)=1-P(z$

From normal distribution table, we will get:

$P(z>1.75)=1-0.95994$

$P(z>1.75)=0.04006$

Therefore, the probability that a randomly selected bill will have an amount greater than $173 would be 0.04006 or 4.001%.

Since z-scores lower than -1.96 or higher than 1.96 are considered unusual. As the bill with size $173 has z-score of 1.75, therefore, a bill with size $173 can not be considered unusual.