Answer:
a) 95%
b) 2%
c) 16%
d) 98%
Explanation:
We have the following:
This is a normal distribution
Mean = 69
Standard Deviation = 6
a) Between 57 and 81%
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ x=57 \\ z=\frac{57-69}{6}=-\frac{12}{6} \\ z=-2 \\ \\ x=81 \\ z=\frac{81-69}{6}=\frac{12}{6} \\ z=2 \\ \end{gathered}[/tex]
The probability that a score is between 57 & 81 is given by the Area between (z = -2) & (z = 2):
[tex]\begin{gathered} P=0.97725-0.02275 \\ P=0.9545 \\ P=95.45\approx95 \\ P=95\text{ \%} \\ \\ \therefore P=95\text{ \%} \end{gathered}[/tex]
b) Above 81%
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ x>81 \\ z=\frac{81-69}{6} \\ z=\frac{12}{6}=2 \\ z=2 \end{gathered}[/tex]
The probability that a score is above 81% is given by the area of the graph greater than (z = 2):
[tex]\begin{gathered} P=0.02275 \\ P=2.275\approx2.3 \\ P=2.3\approx2 \\ P=2\text{ \%} \\ \\ \therefore P=2\text{ \%} \end{gathered}[/tex]
c) Below 63%
[tex]\begin{gathered} x<63 \\ z=\frac{63-69}{6} \\ z=-\frac{6}{6}=-1 \\ z=-1 \end{gathered}[/tex]
The probability that a score is below 63% is given by the area of the graph lesser than (z = -1):
[tex]\begin{gathered} P=0.15866 \\ P=15.866\approx16 \\ P=16\text{ \%} \end{gathered}[/tex]
d) Between 51 and 81
[tex]\begin{gathered} 51\le x\le81 \\ z=\frac{51-69}{6} \\ z=-\frac{18}{6}=-3 \\ z=-3 \\ \\ z=\frac{81-69}{6} \\ z=\frac{12}{6}=2 \\ z=2 \end{gathered}[/tex]
The probability that a score is between 51 & 81 is given by the Area between (z = -3 & (z = 2):
[tex]\begin{gathered} P=0.97725-0.00135 \\ P=0.9759 \\ P=97.59\approx98 \\ P=98\text{ \%} \end{gathered}[/tex]
