Answer:
The percent of households with rates from $100 to $115. is [tex]P(100 < x < 115) =[/tex]94.1%
Step-by-step explanation:
From the question we are told that
The mean rate is [tex]\mu =[/tex]$ 106.50 per month
The standard deviation is [tex]\sigma =[/tex]$3.85
Let the lower rate be [tex]a =[/tex]$100
Let the higher rate be [tex]b =[/tex]$ 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
[tex]P(a < x < b) = P[ \frac{a -\mu}{\sigma } } < \frac{x- \mu}{\sigma} < \frac{b - \mu }{\sigma } ][/tex]
here x is a random value rate which lies between the higher rate and the lower rate so
[tex]P(100 < x < 115) = P[ \frac{100 -106.50}{3.85} } < \frac{x- \mu}{\sigma} < \frac{115 - 106.50 }{3.85 } ][/tex]
[tex]P(100 < x < 115) = P[ -1.688< \frac{x- \mu}{\sigma} < 2.208 ][/tex]
Where
[tex]z = \frac{x- \mu}{\sigma}[/tex]
Where z is the standardized value of x
So
[tex]P(100 < x < 115) = P[ -1.688< z < 2.208 ][/tex]
[tex]P(100 < x < 115) = P(z< 2.208 ) - P(z< -1.69 )[/tex]
Now from the z table we obtain that
[tex]P(100 < x < 115) = 0.9864 - 0.0455[/tex]
[tex]P(100 < x < 115) = 0.941[/tex]
[tex]P(100 < x < 115) =[/tex]94.1%
