We have been given that a set of art exam scores are normally distributed with a mean of 81 points and a standard deviation of 10 points. Kamil got a score of 78 points on the exam. We are asked to find the proportion of exam scores that are lower than Kamil's score.
First of all, we will find z-score corresponding to 78.
[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{78-81}{10}[/tex]
[tex]z=\frac{-3}{10}[/tex]
[tex]z=-0.3[/tex]
Now we will use normal distribution table to find the probability under a z-score of [tex]-0.3[/tex] that is [tex]P(z<-0.3)[/tex].
[tex]P(z<-0.3)=0.38209[/tex]
Upon rounding to 4 decimal places, we will get:
[tex]P(z<-0.3)\approx 0.3821[/tex]
Therefore, [tex]0.3821[/tex] of exam scores are lower than Kamil's score.
Answer:
is 0.3821
Step-by-step explanation:
just answered it on khan
