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?

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.

, where,

Substitute the given values:

Now, we will use formula as:

From normal distribution table, we will get:

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.

is there any attachment? what are the equations?

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!