Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]P(X > 66) = P(Z> 1 ) = 0.15866[/tex]
b
[tex]P(\= X > 66) = P(Z> 2 ) = 0.02275[/tex]
c
[tex]P(\= X > 66) = P(Z> 10 ) = 0[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 64 \ inches[/tex]
The standard deviation is [tex]\sigma = 2 \ inches[/tex]
The probability that a randomly selected woman is taller than 66 inches is mathematically represented as
[tex]P(X > 66) = P(\frac{X - \mu }{\sigma } > \frac{ 66 - \mu }{\sigma} )[/tex]
Generally [tex]\frac{ X - \mu }{\sigma } = Z(The \ standardized \ value \ of \ X )[/tex]
So
[tex]P(X > 66) = P(Z> \frac{ 66 - 64 }{ 2} )[/tex]
[tex]P(X > 66) = P(Z> 1 )[/tex]
From the z-table the value of [tex]P(Z > 1 ) = 0.15866[/tex]
So
[tex]P(X > 66) = P(Z> 1 ) = 0.15866[/tex]
Considering b
sample mean is n = 4
Generally the standard error of mean is mathematically represented as
[tex]\sigma _{\= x} = \frac{\sigma }{\sqrt{4} }[/tex]
=> [tex]\sigma _{\= x} = \frac{2 }{\sqrt{4} }[/tex]
=> [tex]\sigma _{\= x} = 1[/tex]
The probability that the sample mean height is greater than 66 inches
[tex]P(\= X > 66) = P(\frac{X - \mu }{\sigma_{\= x } } > \frac{ 66 - \mu }{\sigma_{\= x }} )[/tex]
=> [tex]P(\= X > 66) = P(Z > \frac{ 66 - 64 }{1} )[/tex]
=> [tex]P(\= X > 66) = P(Z> 2 )[/tex]
From the z-table the value of [tex]P(Z > 2 ) = 0.02275[/tex]
=> [tex]P(\= X > 66) = P(Z> 2 ) = 0.02275[/tex]
Considering b
sample mean is n = 100
Generally the standard error of mean is mathematically represented as
[tex]\sigma _{\= x} = \frac{2 }{\sqrt{100} }[/tex]
=> [tex]\sigma _{\= x} = 0.2[/tex]
The probability that the sample mean height is greater than 66 inches
[tex]P(\= X > 66) = P(Z > \frac{ 66 - 64 }{0.2} )[/tex]
=> [tex]P(\= X > 66) = P(Z> 10 )[/tex]
From the z-table the value of [tex]P(Z > 10 ) = 0[/tex]
[tex]P(\= X > 66) = P(Z> 10 ) = 0[/tex]
