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Alla [95]
3 years ago
14

The monthly utility bills in a city are normally​ distributed, with a mean of ​$ and a standard deviation of ​$. Find the probab

ility 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.)

Mathematics
1 answer:
svetoff [14.1K]3 years ago
5 0

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

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