We have been given that the test scores on a final exam are normally distributed with a mean of 74 and a standard deviation of 3. We are asked to find the probability that a randomly selected test has a score higher than 77.
First of all, we will find z-score corresponding to sample score 77.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
z = z-score,
x = Random sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
[tex]z=\frac{77-74}{3}[/tex]
[tex]z=\frac{3}{3}[/tex]
[tex]z=1[/tex]
Now we need to find [tex]P(z>1)[/tex].
We will use formula [tex]P(z>a)=1-P(z<a)[/tex] to find the probability greater than a z-score of 1.
[tex]P(z>1)=1-P(z<1)[/tex]
Using normal distribution table, we will get:
[tex]P(z>1)=1-0.84134[/tex]
[tex]P(z>1)=0.15866[/tex]
Therefore, the probability that a randomly selected test has a score higher than 77 would be 0.15866.
