34% of the scores lie between 433 and 523.
Solution:
Given data:
Mean (μ) = 433
Standard deviation (σ) = 90
Empirical rule to determine the percent:
(1) About 68% of all the values lie within 1 standard deviation of the mean.
(2) About 95% of all the values lie within 2 standard deviations of the mean.
(3) About 99.7% of all the values lie within 3 standard deviations of the mean.
[tex]$Z(X)=\frac{x-\mu}{\sigma}[/tex]
[tex]$Z(433)=\frac{433-\ 433}{90}=0[/tex]
[tex]$Z(523)=\frac{523-\ 433}{90}=1[/tex]
Z lies between o and 1.
P(433 < x < 523) = P(0 < Z < 1)
μ = 433 and μ + σ = 433 + 90 = 523
Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.
i. e. [tex]((\mu-\sigma) \ \text{to} \ (\mu+\sigma))=68\%[/tex]
Here μ to μ + σ = [tex]\frac{68\%}{2} =34\%[/tex]
Hence 34% of the scores lie between 433 and 523.
