a) ![X](https://tex.z-dn.net/?f=X%3C67.1)
b) 0.1539
c) 0.1539
d) 0.6922
Step-by-step explanation:
a)
In this problem, the score on the exam is normally distributed with the following parameters:
(mean)
(standard deviation)
We call X the name of the variable (the score obtained in the exam).
Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:
![X](https://tex.z-dn.net/?f=X%3C67.1)
And the probability for this to occur can be written as:
![p(X](https://tex.z-dn.net/?f=p%28X%3C67.1%29)
b)
To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between
and
, where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.
The z-score corresponding to 67.1 is:
![Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B67.1-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B67.1-78.1%7D%7B10.8%7D%3D-1.02)
Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:
![p(X](https://tex.z-dn.net/?f=p%28X%3C67.1%29%3Dp%28z%3C-1.02%29)
And by looking at the z-score tables, we find that this probability is:
![p(z](https://tex.z-dn.net/?f=p%28z%3C-1.02%29%3D0.1539)
And so,
![p(X](https://tex.z-dn.net/?f=p%28X%3C67.1%29%3D0.1539)
c)
Here we want to find the probability that a randomly chosen score is greater than 89.1, so
![p(X>89.1)](https://tex.z-dn.net/?f=p%28X%3E89.1%29)
First of all, we have to calculate the z-score corresponding to this value of X, which is:
![Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7B89.1-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B89.1-78.1%7D%7B10.8%7D%3D1.02)
Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:
![p(z>1.02) =p(z](https://tex.z-dn.net/?f=p%28z%3E1.02%29%20%3Dp%28z%3C-1.02%29)
Because the normal distribution is symmetric.
But from part b) we know that
![p(z](https://tex.z-dn.net/?f=p%28z%3C-1.02%29%3D0.1539)
Therefore:
![p(X>89.1)=p(z>1.02)=0.1539](https://tex.z-dn.net/?f=p%28X%3E89.1%29%3Dp%28z%3E1.02%29%3D0.1539)
d)
Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as
![p(67.1](https://tex.z-dn.net/?f=p%2867.1%3CX%3C89.1%29)
Or also as
![p(67.1](https://tex.z-dn.net/?f=p%2867.1%3CX%3C89.1%29%3D1-p%28X%3C67.1%29-p%28X%3E89.1%29)
Since the overall probability under the whole distribution must be 1.
From part b) and c) we know that:
![p(X](https://tex.z-dn.net/?f=p%28X%3C67.1%29%3D0.1539)
![p(X>89.1)=0.1539](https://tex.z-dn.net/?f=p%28X%3E89.1%29%3D0.1539)
Therefore, here we find immediately than:
![p(67.1](https://tex.z-dn.net/?f=p%2867.1%3CX%3C89.1%29%3D1-0.1539-0.1539%3D0.6922)