The average length of time required to complete a college achievement test is approximately normal with a mean of 80 minutes and a standard deviation of 11 minutes. When should the test be terminated (in minutes) if you wish to allow sufficient time for 85% of the students to complete the test? (Round your answer to two decimal places.)
____ min
z value at 85% = 1.04
z = (x - mean)/s
1.04 = (x - 80 )/11
x = 11 *1.04 + 80
x = 91.44
The average length of time required to complete a college achievement test is approximately normal with...
8. (8 pts) The length of time for students to complete a test is found to be normally distributed with mean 70 minutes and standard deviation 12 minutes. How long should the time for the test be if we wish for 90 percent of the students to complete the test in less time than the test length? (Use the Normal table provided for your calculations)
The time needed for college students to complete a certain paper and pencil maze follows a Normal distribution with a population mean of 80 seconds and a population standard deviation of 16 seconds. You wish to see if the mean time µ is changed by meditation, so you have a random sample of 12 college students meditate for 30 minutes and then complete the maze. It takes them an average of x = 73.3 seconds to complete the maze, with...
(1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 3.8 seconds. You wish to see if the mean time u is changed by vigorous exercise, so you have a group of 15 college students randomly selected exercise vigorously for 30 minutes and then complete the maze. It takes them an average of x = 28.8 seconds to complete the maze....
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5. 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. Answer the following questions. What is the probability of completing the exam in one hour or less? (Round to four decimal places) What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes? (Round to four decimal places) Assume that...
The time needed for college students to complete a certain paper
and pencil maze follows a Normal distribution, with a mean of 70
seconds and a standard deviation of 15 seconds. You wish to see if
the mean time μ is changed by meditation, so you have a group of 9
college students meditate for 30 minutes and then complete the
maze.
Question 29 of 40> The time needed for college students to complete a certain paper and pencil maze...
(1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 3.7 seconds. You wish to see if the mean time u is changed by vigorous exercise, so you have a group of 25 college students exercise vigorously for 30 minutes and then complete the maze. It takes them an average of x = 28 seconds to complete the maze. Use this...
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. Answer the following questions. (a) What is the probability of completing the exam in one hour or less? (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. .0233 (b) What is the probability that a student will complete the exam in more than 60 minutes but less than...
The time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes. How much time should be given to complete the exam so that 75% of the students will complete the exam in the time given?