Question

ACT math scores for a particular year are approximately normally distributed

Answer 1

Answer:

$a. \ \ \ P(X>30.4)=0.1587\\\\b. \ \ \ P(X$

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:

$z=\frac{\bar X-\mu}{\sigma}\\\\P(X>30.4)=P(z>\frac{30.4-28}{2.4})\\\\=P(z>1)=1-P(z$

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:

$z=\frac{\bar X-\mu}{\sigma}\\\\P(X$

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:

$P(25.6$

Hence, the probability that a random selection has a score between 25.6 and 32.8 is 0.8186