Answer: 5.37
Step-by-step explanation:
Let x = ACT scores.
Given: ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal.
i.e. P(X>28)=0.09 (i)
Now,
[tex]P(X>28)=P(\dfrac{X-\mu}{\sigma}>\dfrac{28-20.8}{\sigma})\\\\= P(z>\dfrac{7.2}{\sigma})\ \ \ \ [z=\dfrac{X-\mu}{\sigma}][/tex] (ii)
One -tailed z value for p-value of 0.09 =1.3408 [By z-table]
From (i) and (ii)
[tex]\dfrac{7.2}{\sigma}=1.3408\\\\\Rightarrow\ \sigma=\dfrac{7.2}{1.3408}\\\\\Rightarrow\ \sigma=5.37[/tex]
Hence, the standard deviation = 5.37