Question

The monthly utility bills in a city are normally​ distributed, with a mean of ​$ and a standard deviation of ​$. Find the probability that a randomly selected utility bill is​ (a) less than ​$​, ​(b) between ​$ and ​$​, and​ (c) more than ​$. ​(a) The probability that a randomly selected utility bill is less than ​$ is 0.0076. ​(Round to four decimal places as​ needed.) ​(b) The probability that a randomly selected utility bill is between ​$ and ​$ is 0.6859. ​(Round to four decimal places as​ needed.) ​(c) The probability that a randomly selected utility bill is more than ​$ is 0.0764. ​(Round to four decimal places as​ needed.)

Answer 1

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

$P(X < 65) = 0.003467$

b

$P(86 < X < 110 )= 0.63928$

c

$P(X > 120 ) =0.062024$

Step-by-step explanation:

From the question we are told that

 The  mean is  $\mu = \ 100$

 The  standard deviation is  $\sigma = \ 13$

 Generally the probability that the  monthly utility bills is less than  $65  is  mathematically represented as

     $P(X < 65) = P(\frac{X - \mu }{\sigma } < \frac{65 - 100 }{13 } )$

Generally

  $\frac{X - \mu }{\sigma } = Z (The\ standardized \ value \ of\ X )$

So

 $P(X < 65) = P(Z < -2.70 )$

From the z-table

    $P(Z < -2.70 ) = 0.003467$

So  

   $P(X < 65) = 0.003467$

Generally the  probability that a randomly selected utility bill is between $86​ and $​110 is mathematically represented as

 $P(86 < X< 110 )= P( \frac{86 -100}{ 13} < \frac{X - \mu}{\sigma } < \frac{86 -100}{ 13} )$

=>  $P(86 < X < 110 )= P( \frac{86 -100}{ 13} < Z < \frac{110 -100}{ 13} )$      

=>    $P(86 < X < 110 )= P(-1.08 < Z < 0.77 )$      

=>    $P(86 < X < 110 )= P(Z < 0.77 ) - P(Z < -1.08 )$    

From the z-table

    $P(Z < 0.77 ) = 0.77935$

and  

   $P(Z < -1.08 ) = 0.14007$

So

     $P(86 < X < 110 )= 0.77935 - 0.14007$    

=>   $P(86 < X < 110 )= 0.63928$    

 Generally the probability that the  monthly utility bills is  more than $120  is mathematically represented as

         $P(X >120 ) = P(\frac{X - \mu }{\sigma } > \frac{120 - 100 }{13 })$

=>   $P(X > 120 ) = P(Z > 1.538 )$

From the z-table

    $P(Z > 1.54 ) = 0.062024$

So  

   $P(X > 120 ) =0.062024$