Complete Question
The mean height of women in a country (ages 20-29) is 64.4 inches. A random sample of 75 women in this age ground is selected. what is the probability that the mean height for the sample is greater than 65 inches? assume [tex]\sigma = 2.97[/tex]
Answer:
The value is [tex]P(X > 65) = 0.039715[/tex]
Step-by-step explanation:
From question we are told that
The mean is [tex]\mu = 64.4 \ inches[/tex]
The sample size is [tex]n = 75[/tex]
The probability that the mean height for the sample is greater than 65 inches is mathematically represented as
[tex]P(X > 65) = P[\frac{X - \mu }{ \sigma_{\= x} } > \frac{65 - 64.4 }{ \sigma_{\= x} } ][/tex]
Where [tex]\sigma _{\= x }[/tex] is the standard error of mean which is evaluated as
[tex]\sigma_{\= x } = \frac{\sigma}{\sqrt{n} }[/tex]
=> [tex]\sigma_{\= x } = \frac{2.97}{\sqrt{75} }[/tex]
=> [tex]\sigma_{\= x } = 0.343[/tex]
Generally [tex]\frac{X - \mu }{ \sigma_{\= x } } = Z(The \ standardized \ value \ of \ X )[/tex]
[tex]P(X > 65) = P[Z> \frac{65 - 64.4 }{0.342 } ][/tex]
So
[tex]P(X > 65) = P[Z >1.754 ][/tex]
From the z-table the value of
[tex]P(X > 65) = P[Z >1.754 ] = 0.039715[/tex]
[tex]P(X > 65) = 0.039715[/tex]
