Statistic student return and forgot it all

[yeungmei]

I studied stat couple years ago, and going back again...here goes...

THe time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes.

Assume that the class has students and that the examination period is 90 mintues in length. How many students do you expect will be unable to complete the exam in the allotted time?

so, i know mean is 80 and SD is 10, and the z is 90...am i find the "n" in the case?? How do i find it? by using what formula??

 

[e(ho0n3]

I think you have to find out what percent of the normal distribution is beyond one standard deviation. Then, multiply that percentage by the amount of students taking the test and you should have your answer.

 

[Architeuthis Dux]

As a fellow scientist, it is understandable that you may have forgotten some of the concepts from your previous study of statistics. However, it is important to refresh your knowledge and skills in order to effectively analyze and interpret data. In this case, we can use the normal distribution formula to estimate the number of students who may not be able to complete the exam in the given time frame. First, we can convert the given values into z-scores by using the formula z = (x - mean)/standard deviation. In this case, the z-score for 90 minutes is (90-80)/10 = 1. This means that the examination period of 90 minutes is one standard deviation above the mean of 80 minutes. From the standard normal distribution table, we can see that the probability of a z-score being greater than 1 is approximately 0.1587. This means that approximately 15.87% of the class may not be able to complete the exam in the allotted time. To find the actual number of students, we can multiply this probability by the total number of students in the class. Therefore, we can expect that approximately 15.87% of the class, or about 16 students, may not be able to complete the exam in the given time frame. I hope this helps refresh your understanding of statistics and its applications. Good luck on your exam!