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.)

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

[tex]P(X < 65) =  0.003467[/tex]

b

[tex]P(86 <  X <  110 )= 0.63928 [/tex]

c

[tex]P(X > 120 ) =0.062024[/tex]

Step-by-step explanation:

From the question we are told that

 The  mean is  [tex]\mu  =  \$ 100[/tex]

 The  standard deviation is  [tex]\sigma  =  \$ 13[/tex]

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

     [tex]P(X < 65) =  P(\frac{X  -  \mu }{\sigma }  < \frac{65  -  100 }{13 }   )[/tex]

Generally

  [tex]\frac{X  -  \mu }{\sigma } =  Z (The\ standardized \ value \ of\  X )[/tex]

So

 [tex]P(X < 65) =  P(Z < -2.70  )[/tex]

From the z-table

    [tex]P(Z < -2.70  ) =  0.003467[/tex]

So  

   [tex]P(X < 65) =  0.003467[/tex]

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

 [tex]P(86 <  X<  110 )=  P( \frac{86 -100}{ 13} <  \frac{X - \mu}{\sigma } < \frac{86 -100}{ 13}    )[/tex]

=>  [tex]P(86 <  X <  110 )=  P( \frac{86 -100}{ 13} < Z < \frac{110 -100}{ 13} )[/tex]      

=>    [tex]P(86 <  X <  110 )=  P(-1.08 < Z < 0.77 )[/tex]      

=>    [tex]P(86 <  X <  110 )= P(Z < 0.77 ) -  P(Z < -1.08 ) [/tex]    

From the z-table

    [tex]P(Z < 0.77 ) = 0.77935[/tex]

and  

   [tex]P(Z < -1.08 ) =  0.14007[/tex]

So

     [tex]P(86 <  X <  110 )= 0.77935  -  0.14007 [/tex]    

=>   [tex]P(86 <  X <  110 )= 0.63928 [/tex]    

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

         [tex]P(X >120 ) =  P(\frac{X  -  \mu }{\sigma }  > \frac{120  -  100 }{13 })[/tex]

=>   [tex]P(X > 120 ) =  P(Z > 1.538   )[/tex]

From the z-table

    [tex]P(Z > 1.54  ) =  0.062024[/tex]

So  

   [tex]P(X > 120 ) =0.062024[/tex]

Can anyone answer these questions?

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