Answer:
The value is [tex]P( X > 80 ) = 0.58901[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 24
The mean is [tex]\mu = 82[/tex]
The standard deviation is [tex]\sigma = 8.9[/tex]
Generally the probability that a randomly chosen student earned a B or higher on the test (score of 80 or higher ) is mathematically represented as
[tex]P( X > 80 ) = P( \frac{ X - \mu }{\sigma } > \frac{ 80 - 82 }{ 8.9 } )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P( X > 80 ) = P(Z > -0.225 )[/tex]
From the z table the area under the normal curve to the right corresponding to -0.225 is
[tex]P(Z > -0.225 ) = 0.58901[/tex]
=> [tex]P( X > 80 ) = 0.58901[/tex]
