Answer:
The percent of households with rates from $100 to $115. is
94.1%
Step-by-step explanation:
From the question we are told that
The mean rate is
$ 106.50 per month
The standard deviation is
$3.85
Let the lower rate be
$100
Let the higher rate be
$ 115
Assumed from the question that the data set is normally
The estimate of the percent of households with rates from $100 to $115. is mathematically represented as
![P(a < x < b) = P[ \frac{a -\mu}{\sigma } } < \frac{x- \mu}{\sigma} < \frac{b - \mu }{\sigma } ]](https://tex.z-dn.net/?f=P%28a%20%3C%20x%20%3C%20b%29%20%3D%20%20P%5B%20%5Cfrac%7Ba%20-%5Cmu%7D%7B%5Csigma%20%7D%20%7D%20%3C%20%20%5Cfrac%7Bx-%20%5Cmu%7D%7B%5Csigma%7D%20%3C%20%5Cfrac%7Bb%20-%20%5Cmu%20%7D%7B%5Csigma%20%7D%20%20%5D)
here x is a random value rate which lies between the higher rate and the lower rate so
![P(100 < x < 115) = P[ \frac{100 -106.50}{3.85} } < \frac{x- \mu}{\sigma} < \frac{115 - 106.50 }{3.85 } ]](https://tex.z-dn.net/?f=P%28100%20%3C%20x%20%3C%20115%29%20%3D%20%20P%5B%20%5Cfrac%7B100%20-106.50%7D%7B3.85%7D%20%7D%20%3C%20%20%5Cfrac%7Bx-%20%5Cmu%7D%7B%5Csigma%7D%20%3C%20%5Cfrac%7B115%20-%20106.50%20%7D%7B3.85%20%7D%20%20%5D)
![P(100 < x < 115) = P[ -1.688< \frac{x- \mu}{\sigma} < 2.208 ]](https://tex.z-dn.net/?f=P%28100%20%3C%20x%20%3C%20115%29%20%3D%20%20P%5B%20-1.688%3C%20%20%5Cfrac%7Bx-%20%5Cmu%7D%7B%5Csigma%7D%20%3C%202.208%20%5D)
Where

Where z is the standardized value of x
So
![P(100 < x < 115) = P[ -1.688< z < 2.208 ]](https://tex.z-dn.net/?f=P%28100%20%3C%20x%20%3C%20115%29%20%3D%20%20P%5B%20-1.688%3C%20z%20%3C%202.208%20%5D)

Now from the z table we obtain that


94.1%