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 the mean is 28 and the standard deviation is 2.4
-Let X be the score of any random selection from the population.
-The probability that a randomly selected score is greater than 30.4:
[tex]z=\frac{\bar X-\mu}{\sigma}\\\\P(X>30.4)=P(z>\frac{30.4-28}{2.4})\\\\=P(z>1)=1-P(z<1)\\\\=1-0.84134\\\\=0.1587[/tex]
Hence, the probability that the score is greater than 30.4 is 0.1587
b. The probability that a randomly selected score is less than 32.8 given mean =28 and the standard deviation is 2.4:
[tex]z=\frac{\bar X-\mu}{\sigma}\\\\P(X<32.8)=P(z<\frac{32.8-28}{2.4})\\\\=0.9773\\\\[/tex]
Hence, the probability that a random score is less than 32.8 is 0.9773
c. The probability that a random selection has a score of between 25.6 and 32.8.
-Given the mean=28 and sd=2.4, the probability is calculated as follows:
[tex]P(25.6<X<32.8)=P(\frac{25.6-28}{2.4}<X<\frac{32.8-28}{2.4})\\\\\\=P(-1<z<2)\\\\=0.9773-0.1587\\\\=0.8186[/tex]
Hence, the probability that a random selection has a score between 25.6 and 32.8 is 0.8186
