Question

The mean rate for cable with Internet from a sample of households was $106.50 per month with a standard deviation of $3.85 per month. Assuming the data set has a normal distribution, estimate the percent of households with rates from $100 to $115.

Answer 1

The percent of households with rates from $100 to $115. is
`P(100 < x < 115) =`94.1%

Explanation:

From the question we are told that

The mean rate is
`\mu =`$ 106.50 per month

The standard deviation is
`\sigma =`$3.85

Let the lower rate be
`a =`$100

Let the higher rate be
`b =`$ 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[ (a -\mu)/(\sigma ) } < (x- \mu)/(\sigma) < (b - \mu )/(\sigma ) ]`

here x is a random value rate which lies between the higher rate and the lower rate so

`P(100 < x < 115) = P[ (100 -106.50)/(3.85) } < (x- \mu)/(\sigma) < (115 - 106.50 )/(3.85 ) ]`

`P(100 < x < 115) = P[ -1.688< (x- \mu)/(\sigma) < 2.208 ]`

Where

`z = (x- \mu)/(\sigma)`

Where z is the standardized value of x

So

`P(100 < x < 115) = P[ -1.688< z < 2.208 ]`

`P(100 < x < 115) = P(z< 2.208 ) - P(z< -1.69 )`

Now from the z table we obtain that

`P(100 < x < 115) = 0.9864 - 0.0455`

`P(100 < x < 115) = 0.941`

`P(100 < x < 115) =`94.1%