Answer:
6.68% of the female college-bound high school seniors had scores above 575.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 50
We are given that the distribution of score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(scores above 575)
P(x > 575)
[tex]P( x > 575) = P( z > \displaystyle\frac{575 - 500}{50}) = P(z > 1.5)[/tex]
[tex]= 1 - P(z \leq 1.5)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 575) = 1 - 0.9332= 0.0668 = 6.68\%[/tex]
6.68% of the female college-bound high school seniors had scores above 575.
