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.

I have to solve using the empirical rule

Can somebody help me please? These are really complicated

A: 16%

B: 95%

C: 97.5%

Step-by-step explanation:

According to the empirical rule:

68% of a normal distribution is between -1 and +1 standard deviations.

95% of a normal distribution is between -2 and +2 standard deviations.

99.7% of a normal distribution is between -3 and +3 standard deviations.

Given μ = 27 and σ = 2.

Part A

29 is one standard deviation above the mean.  We can show this by calculating the z-score:

z = (x − μ) / σ

z = (29 − 27) / 2

z = 1

We know that 68% is between -1 and +1 standard deviations.  Since normal distributions are symmetrical, we can also say that 34% is between 0 and +1 standard deviations.

P(0 < Z < 1) = 68%/2 = 34%

We can also say that 50% is less than 0 standard deviations.

P(Z < 0) = 50%

Therefore, P(Z < 1) = 34% + 50% = 84%.

Which means P(Z > 1) = 100% − 84% = 16%.

Part B

Like before, calculate the z-scores:

z₁ = (23 − 27) / 2

z₁ = -2

z₂ = (31 − 27) / 2

z₂ = 2

From the empirical rule, we know this is 95% of the normal distribution.

Part C

We found in part B that the z-score is 2.

P(0 < Z < 2) = 95%/2 = 47.5%

P(Z < 2) = 50% + 47.5% = 97.5%

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: