Answer:
The probability is [tex]P( \= X > 268 ) =0.35376[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 266[/tex]
The standard deviation is [tex]\sigma = 16[/tex]
Generally the standard error of mean is mathematically represented as
[tex]\sigma_{\= x} = \frac{\sigma}{\sqrt{n} }[/tex]
=> [tex]\sigma_{\= x} = \frac{16}{\sqrt{9} }[/tex]
=> [tex]\sigma_{\= x} = 5.33[/tex]
Generally the probability that the average pregnancy length for nine randomly chosen women exceeds 268 days is mathematically represented as
[tex]P( \= X > 268 ) = P (\frac{ \= x - \mu }{ \sigma_{\= x}} > \frac{268 - 266}{5.33 } )[/tex]
[tex]\frac{\= X -\mu}{\sigma_{\= x} } = Z (The \ standardized \ value\ of \ \= X )[/tex]
[tex]P( \= X > 268 ) = P (Z > 0.3752 )[/tex]
From the z table the area under the normal curve to the right corresponding to 0.3752 is
P (Z > 0.3752) = 0.35376
So
[tex]P( \= X > 268 ) =0.35376[/tex]
