Answer:
The probability is [tex]P(\= X > 103.7 ) = 0.9862 [/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 106 \ gallons[/tex]
The population standard deviation is [tex]\sigma = 6 \ gallons[/tex]
The sample size is [tex]n = 33[/tex]
Generally the standard deviation of sample mean is mathematically represented as
[tex]\sigma_{x} = \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]\sigma_{x} = \frac{6}{\sqrt{33} }[/tex]
=> [tex]\sigma_{x} = 1.044[/tex]
Generally the probability that the sample mean would be greater than 103.7 gallons is mathematically represented as
[tex]P(\= X > 103.7 ) = P(\frac{X - \mu }{\sigma_{x}} > \frac{103.7 - 106}{1.044 } )[/tex]
Generally [tex]\frac{X - \mu }{\sigma_{x}} = Z (The \ standardized \ value \ of \= X )[/tex]
So
[tex]P(\= X > 103.7 ) = P(Z > -2.203 )[/tex]
From the z table the probability of (Z > -2.203 ) is
[tex]P(Z > -2.203 ) = 0.9862[/tex]
So
[tex]P(\= X > 103.7 ) = 0.9862 [/tex]
