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]
