the average length of time required for students to complete a test in a an introductory statistics course is found to be 50 minutes with a standard deviation of 12 minutes. if students are allowed 80 minutes to complete the test , approximately what percentage of the students will have sufficient time to complete the test? ( assume that the time required to complete the test follows a normal distribution)
To find the percentage of students who will have sufficient time to complete the test, we need to calculate the area under the normal distribution curve up to the value of 80 minutes. This can be done by identifying the z-score for 80 minutes and referring to the standard normal distribution table.
The z-score is calculated using the formula:
z = (x - μ) / σ
Where:
x = value you want to find the z-score for (80 minutes)
μ = population mean (50 minutes)
σ = population standard deviation (12 minutes)
Plugging in the values:
z = (80 - 50) / 12 = 2.5
Now we need to find the area under the standard normal distribution curve up to the z-score of 2.5.
Using the standard normal distribution table, we can find that the area to the left of the z-score 2.5 is approximately 0.9938.
To convert this to a percentage, we multiply by 100:
0.9938 * 100 = 99.38%
Therefore, approximately 99.38% of the students will have sufficient time to complete the test.