Answer:
[tex]a. \ \ \ P(X>30.4)=0.1587\\\\b. \ \ \ P(X<32.8)=0.9773\\\\c.\ \ \ P(25.6<x<32.8)=0.8186[/tex]
Step-by-step explanation:
a. Given that the mean is 28 and the standard deviation is 2.4.
Let X be the score of our random selection.
-The z-score is calculated using the formula:
[tex]z=\frac{\bar X-\mu}{\sigma}[/tex]
-The probability that the score is greater than 30.4 is calculated as:
[tex]P(X>30.4)=P(z<\frac{\bar X-\mu}{\sigma})[/tex]
[tex]P(X>)=P(z>\frac{\bar X-\mu}{\sigma})\\\\z=\frac{30.4-28}{2.4}=1.0\\\\\therefore P(z>1)=1-0.84134\\\\=0.1587[/tex]
Hence, the probability of a random score being greater than 30.4 is 0.1587
b. The probability that a randomly selected score is less than 32.8:
Mean=28, sd=2.4 and X=32.8
[tex]P(X<32.8)=P(z<\frac{\bar X-\mu}{\sigma})\\\\=P(z<\frac{32.8-28}{2.4})\\\\=P(z<2.0)=0.9773[/tex]
Hence, the probability that a random selection has a score less than 32.8 is 0.9773
c. The probability that a randomly selected score is between 25.6 and 32.8:
mean=28, sd=2.4 X=25.6, 32.8
[tex]P(25.6<X<32.8)=P(\frac{25.6-28}{2.4}<z<\frac{32.8-28}{2.4}\\\\=P(-1<z<2.0)\\\\=0.9773-0.1587\\\\=0.8186[/tex]
Hence, the probability of the score being between 25.6 and 32.8 is 0.8186
