Answer:
[tex]P(x < 535.8) = 0.64[/tex]
[tex]P_{64} = 535.8[/tex]
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 100
We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.64
P(X<x) = 0.64
[tex]P( X < x) = P( z < \displaystyle\frac{x - 500}{10})=0.64[/tex]
Calculation the value from standard normal z table, we have, [tex]p(z<0.358) = 0.64[/tex]
[tex]\displaystyle\frac{x - 500}{100} = 0.358\\x = 535.8[/tex]
[tex]P(x < 535.8) = 0.64[/tex]
[tex]P_{64} = 535.8[/tex]