Question

Final exam scores are normally distributed with a mean of 74 and a standard deviation of 6. Approximately, what percentage of final exam scores would be between 68 and 86?

Answer 1

Answer:

81.86%

Step-by-step explanation:

We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.

First of all we will find z-score using z-score formula.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{68-74}{6}$

$z=\frac{-6}{6}=-1$

Now let us find z-score for 86.

$z=\frac{86-74}{6}$    

$z=\frac{12}{6}=2$        

Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.

Using normal distribution table we will get,    

$P(-1<Z)= .15866$

$P(2>Z)=.97725$  

We will use formula $P(a<Z<b) = P(Z<b) - P(Z<a)$ to find the probability to find that a normal variable lies between two values.

Upon substituting our given values in above formula we will get,

$P(-1<Z<2) = P(Z<2) - P(Z<-1)$

$P(-1<Z<2) = 0.97725-0.15866=0.81859$

Upon converting 0.81859 to percentage we will get

$0.81859*100=81.859\approx 81.86$

Therefore, 81.86% of final exam score will be between 68 and 86.