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

Answer:

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

Step-by-step 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

         $P(a < x < b) = P[ \frac{a -\mu}{\sigma } } < \frac{x- \mu}{\sigma} < \frac{b - \mu }{\sigma } ]$

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

     $P(100 < x < 115) = P[ \frac{100 -106.50}{3.85} } < \frac{x- \mu}{\sigma} < \frac{115 - 106.50 }{3.85 } ]$

      $P(100 < x < 115) = P[ -1.688< \frac{x- \mu}{\sigma} < 2.208 ]$

Where  

      $z = \frac{x- \mu}{\sigma}$

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%