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:
is the mean of the distribution
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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B30.4-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B30.4-28%7D%7B2.4%7D%3D1)
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](https://tex.z-dn.net/?f=p%28z%3EZ%29%20%3D%20p%28z%3C-Z%29)
Here we want to find
, which is therefore equivalent to
. Looking at the z-tables, we find that
![p(z](https://tex.z-dn.net/?f=p%28z%3C-1%29%3D0.1587)
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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B32.8-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B32.8-28%7D%7B2.4%7D%3D2)
Now we notice that:
(1)
Since the overall probability under the curve must be 1. We also note that (from part A)
![p(z>Z) = p(z](https://tex.z-dn.net/?f=p%28z%3EZ%29%20%3D%20p%28z%3C-Z%29)
Which means that we can rewrite (1) as
![p(z](https://tex.z-dn.net/?f=p%28z%3CZ%29%20%3D%201-p%28z%3C-Z%29)
Here, we have
Z = 2
This means that
![p(z](https://tex.z-dn.net/?f=p%28z%3C2%29%3D1-p%28z%3C-2%29)
Looking at the z-tables, we find that
![p(z](https://tex.z-dn.net/?f=p%28z%3C-2%29%3D0.0228)
Therefore, we get
![p(z](https://tex.z-dn.net/?f=p%28z%3C2%29%3D1-0.0228%3D0.9772)
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](https://tex.z-dn.net/?f=z_1%3D%5Cfrac%7B25.6-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B25.6-28%7D%7B2.4%7D%3D-1)
![z_2=\frac{32.8-\mu}{\sigma}=\frac{32.8-28}{2.4}=2](https://tex.z-dn.net/?f=z_2%3D%5Cfrac%7B32.8-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B32.8-28%7D%7B2.4%7D%3D2)
So here we basically want to find the probability that
![p(z_1](https://tex.z-dn.net/?f=p%28z_1%20%3Cz%3Cz_2%29)
Which can be rewritten as:
![p(z_1](https://tex.z-dn.net/?f=p%28z_1%3Cz%3Cz_2%29%3D1-p%28z%3Cz_1%29-p%28z%3Ez_2%29)
So in this case,
![p(-1](https://tex.z-dn.net/?f=p%28-1%3Cz%3C2%29%3D1-p%28z%3C-1%29%20-p%28z%3E2%29)
From part A and B we found that:
![p(z](https://tex.z-dn.net/?f=p%28z%3C-1%29%3D0.1587)
![p(z>2)=1-p(z](https://tex.z-dn.net/?f=p%28z%3E2%29%3D1-p%28z%3C2%29%3D1-0.9772%3D0.0228)
Therefore,
![p(-1](https://tex.z-dn.net/?f=p%28-1%3Cz%3C2%29%3D1-0.1587-0.0228%3D0.8185)