historically, the mean timr to complete this statistics test is 54 minutes with a standard deviation of 9 minutes. how much time should i allow, if my goal is for 95% of the students to be able to complete this test
To determine how much time you should allow for 95% of the students to be able to complete the test, you can use the concept of Z-scores and the standard normal distribution.
The given mean time to complete the test is 54 minutes, and the standard deviation is 9 minutes.
We want to find the amount of time, T, such that 95% of the students will complete the test within that time.
First, we need to determine the Z-score corresponding to the desired percentage. To find the Z-score, we can use a standard normal distribution table or a statistical calculator.
Since we want 95% of the students to complete the test, this means that 5% will take longer than the desired time. Since the standard normal distribution is symmetric, we can look up the Z-score corresponding to the middle 2.5% (0.975) on the positive side of the distribution.
Using a Z-table or calculator, we find that the Z-score corresponding to 0.975 is approximately 1.96.
Now, we can use the formula for Z-score:
Z = (T - μ) / σ
Where Z is the Z-score, T is the desired time, μ is the mean time, and σ is the standard deviation.
Rearranging the formula to solve for T:
T = Z * σ + μ
Plugging in the values:
T = 1.96 * 9 + 54
T ≈ 17.64 + 54
T ≈ 71.64
Therefore, you should allow approximately 71.64 minutes for 95% of the students to be able to complete the test.