Respuesta :
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 :
[tex]P(Z=\frac{X-\mu}{\sigma}>\frac{a-\mu}{\sigma})=P(Z>\frac{173-108.43}{36.98})\\\\\implies P(Z>1.75)[/tex]
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:
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.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{Z-score}[/tex]
[tex]x=\text{Random sample score}[/tex]
[tex]\mu=\text{Mean}[/tex]
[tex]\sigma=\text{Standard deviation}[/tex]
Substitute the given values:
[tex]z=\frac{173-108.43}{36.98}[/tex]
[tex]z=\frac{64.57}{36.98}[/tex]
[tex]z=1.75[/tex]
Now, we will use formula [tex]P(z>a)=1-P(z<a)[/tex] as:
[tex]P(z>1.75)=1-P(z<1.75)[/tex]
From normal distribution table, we will get:
[tex]P(z>1.75)=1-0.95994[/tex]
[tex]P(z>1.75)=0.04006[/tex]
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.
