The time needed to complete a final examination in a particular college course is normally distributed with a mean of 79 minutes and a standard deviation of 8 minutes. Answer the following questions. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time (to 2 decimals)?

z-score for you time of 90 minutes

= (90-79)/8 = 1.375

from tables or from my favourite webpage
probability not finishing = .0846

so of 60 students not finishing
= .0846(60)
= 5.076

so 5 students will not finish

http://davidmlane.com/hyperstat/z_table.html

time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes. Using the 68-95-99.7 rule, if students are given 90 minutes to complete the exam, what percent of students will not finish?

A researcher is interested in manual dexterity of a group of adults, part of study involves asking the participants to complete a task and you a presented with the following information about the length of time taken to complete it in minutes

Mean=20
Sd=4
Assuming that the data is normally distributed
What does it tell about a participant that took 25 minutes to complete the task
If the researcher collected data from 1000 participants, how many people are likely to have completed the task within 16 minutes

To find out how many students are expected to be unable to complete the exam in the allotted time, we need to calculate the proportion of students whose exam time exceeds the allotted time.

First, we need to find the z-score for the allotted time of 90 minutes using the formula:

z = (x - μ) / σ

where x is the allotted time, μ is the mean, and σ is the standard deviation.

Plugging in the values, we get:

z = (90 - 79) / 8 = 1.375

Next, we need to find the area under the normal distribution curve to the right of this z-score. This area represents the proportion of students who will take more time than the allotted 90 minutes.

Using a standard normal distribution table or a calculator, we can find that the area to the right of 1.375 is approximately 0.0848.

To find out the number of students who will be unable to complete the exam in the allotted time, we multiply the proportion by the total number of students in the class:

Number of students unable to complete the exam = 0.0848 * 60 = 5.088

Rounding to two decimal places, we expect approximately 5.09 students to be unable to complete the exam in the allotted time.