Question

ACT mathematics score for a particular year are normally distributed with a mean of 28 and a standard deviation of 2.4 points

Answer 1

A) 0.1587

B) 0.9772

C) 0.8185

Step-by-step explanation:

A)

In this problem, the mathematics score of the year is distributed according to a normal distribution, with parameters:

$\mu=28$ is the mean of the distribution

$\sigma = 2.4$ is the standard deviation of the distribution

We want to find the probability that a randomly selected score is greater than 30.4. First of all, we calculated the z-score associated to this value, which is given by:

$z=\frac{30.4-\mu}{\sigma}=\frac{30.4-28}{2.4}=1$

The z-score tables give the probability that the z-score is less than a certain value; since the distribution is symmetrical around 0,

$p(z>Z) = p(z$

Here we want to find $p(z>1)$, which is therefore equivalent to $p(z$. Looking at the z-tables, we find that

$p(z$

B)

Here instead we want to find the probability that a randomly selected score is less than 32.8.

First of all, we calculate again the z-score associated to this value:

$z=\frac{32.8-\mu}{\sigma}=\frac{32.8-28}{2.4}=2$

Now we notice that:

$p(zZ)$ (1)

Since the overall probability under the curve must be 1. We also note that (from part A)

$p(z>Z) = p(z$

Which means that we can rewrite (1) as

$p(z$

Here, we have

Z = 2

This means that

$p(z$

Looking at the z-tables, we find that

$p(z$

Therefore, we get

$p(z$

C)

Here we want to find the probablity that the score is between 25.6 and 32.8.

First of all, we calculate the z-scores associated to these two values:

$z_1=\frac{25.6-\mu}{\sigma}=\frac{25.6-28}{2.4}=-1$

$z_2=\frac{32.8-\mu}{\sigma}=\frac{32.8-28}{2.4}=2$

So here we basically want to find the probability that

$p(z_1$

Which can be rewritten as:

$p(z_1$

So in this case,

$p(-1$

From part A and B we found that:

$p(z$

$p(z>2)=1-p(z$

Therefore,

$p(-1$