Given the scores on a statewide standardized test are normally distributed
Mean = μ = 78
Standard deviation = σ = 3
Normalize the data using the z-score by using the following formula and chart:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Estimate the percentage of scores of the following cases:
(a) between 75 and 81
so, the z-score for the given numbers will be:
[tex]\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=\frac{-3}{3}=-1 \\ 81\rightarrow z=\frac{81-78}{3}=\frac{3}{3}=1 \end{gathered}[/tex]
As shown, the percentage when (-1 < z < 1) = 68%
(b) above 87
[tex]87\rightarrow z=\frac{87-78}{3}=\frac{9}{3}=3[/tex]
The percentage when (z > 3) = 0.5%
(c) below 72
[tex]72\rightarrow z=\frac{72-78}{3}=\frac{-6}{3}=-2[/tex]
The percentage when (z < -2) = 0.5 + 2 = 2.5%
(d) between 75 and 84
[tex]\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=-\frac{3}{3}=-1 \\ 84\rightarrow z=\frac{84-78}{3}=\frac{6}{3}=2 \end{gathered}[/tex]
The percentage when ( -1 < z < 2 ) = 68 + 13.5 = 81.5%