Answer:
The probability that a randomly selected bill will have an amount greater than $155 is 0.07353.
No, a bill with $155 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 $32.09.
Let us find the z-score for the raw score 155, using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{ z-score}[/tex],
[tex]x=\text{ Raw score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
Upon substituting our given values in above formula we will get,
[tex]z=\frac{155-108.43}{32.09}[/tex]
[tex]z=\frac{46.57}{32.09}[/tex]
[tex]z=1.4512\approx 1.45[/tex]
Let us find probability of z-score greater than 1.45.
[tex]p(z>1.45)=1-P(z<1.45)[/tex]
Using normal distribution table we will get,
[tex]p(z>1.45)=1-0.92647 [/tex]
[tex]p(z>1.45)=0.07353[/tex]
Therefore, the probability that a randomly selected bill will have an amount greater than $155 is 0.07353 or approximately 7.35%.
Since z-scores lower than -1.96 or higher than 1.96 are considered unusual. As the bill with size $155 has z-score of 1.45, therefore, a bill with size $155 can not be considered unusual.
