Answer:
The value is [tex]P( \= X \ge 99) = 0.69146[/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 = 10[/tex]
The sample size is n = 25
The standard deviation of the sampling distribution is
[tex]\sigma_{\= x} = \frac{\sigma}{\sqrt{n} }[/tex]
=> [tex]\sigma_{\= x} = \frac{10}{\sqrt{25} }[/tex]
=> [tex]\sigma_{\= x} = \frac{10}{5 }[/tex]
=> [tex]\sigma_{\= x} = 2[/tex]
Generally the probability of randomly sampling and finding a sample mean of 99 or more is mathematically represented as
[tex]P( \= X \ge 99) = 1 - P(\= X < 99)[/tex]
Here
[tex]P(\= X < 99) = P(\frac{\= X - \mu }{\sigma } < \frac{99 - 100 }{2} )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(\= X < 99) = P(Z < -0.5 )[/tex]
From the z table probability of (Z < -0.5 ) is
[tex]P(Z < -0.5 ) = 0.30854[/tex]
Generally
[tex]P( \= X \ge 99) = 1 - 0.30854[/tex]
=> [tex]P( \= X \ge 99) = 0.69146[/tex]
