Answer :
Since the scores are normally distributed, we can use standard normal distribution to answer this question
First we need to convert the scores 396 and 632 in z scores.
Mean scores = 514
Standard deviation = 118
396 converted to z score will be:
[tex]\frac{396-514}{118} =-1[/tex]
632 converted to z score will be:
[tex] \frac{632-514}{114} =1[/tex]
Using the z-table, we are to find the percentage of scores that is between -1 and 1. The value comes to be 0.68 or 68%
So, expressed to nearest whole percent, 68% of students received score between 396 and 632.
First we need to convert the scores 396 and 632 in z scores.
Mean scores = 514
Standard deviation = 118
396 converted to z score will be:
[tex]\frac{396-514}{118} =-1[/tex]
632 converted to z score will be:
[tex] \frac{632-514}{114} =1[/tex]
Using the z-table, we are to find the percentage of scores that is between -1 and 1. The value comes to be 0.68 or 68%
So, expressed to nearest whole percent, 68% of students received score between 396 and 632.