Question

ACT mathematics score for a particular year are normally distributed with a mean of 27 and a standard deviation of 2 points

Answer 1

Answer:

A) the probability that a randomly selected score is greater than 29 points is 0.1587

B) The percentage of students scores are between 31 and 23 is 95.44%

C) A student who scores 31 is in the 97.72% percentile

Step-by-step explanation:

ACT mathematics score for a particular year are normally distributed with a mean of 27 and a standard deviation of 2 points.

Part A: What is the probability that a randomly selected score is greater than 29 points?

Part B: What percentage of students scores are between 31 and 23?  

Part C: A student who scores 31 is in the ______ percentile.

A) Given that:

Mean (m) = 27 and standard deviation (s) = 2 points

Since the ACT mathematics score is normally distributed, we can use z score. To calculate Z score we use the equation:

$Z=\frac{x-m}{s}$

substituting values:

$Z=\frac{x-m}{s}==\frac{29-27}{2}=1\\Z=1$

P(X > 29) = P(Z > 1) = 1 - P(Z<1)

Using Z tables

P(X > 29) = P(Z > 1) = 1 - P(Z<1) = 1 - 0.8413 = 0.1587 = 15.87%

P(X > 29) = 0.1587

the probability that a randomly selected score is greater than 29 points is 0.1587

B) For score of 31

$Z=\frac{x-m}{s}==\frac{31-27}{2}=2\\Z=2$

For score of 23

$Z=\frac{x-m}{s}==\frac{23-27}{2}=-2\\Z=-2$

P(23 < X < 31) = P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 -  0.0228 = 0.9544

P(23 < X < 31) = 0.9544 = 95.44%

The percentage of students scores are between 31 and 23 is 95.44%

C) For score of 31

$Z=\frac{x-m}{s}==\frac{31-27}{2}=2\\Z=2$

P(X < 31) = P(Z < 2) = 0.9772 = 97.72%

A student who scores 31 is in the 97.72% percentile