If 1,000 students take a test that has a mean of 40 minutes, a standard deviation of 8 minutes, and is normally distributed, how many would you expect would finish in between 24 and 48 minutes?
http://davidmlane.com/hyperstat/z_table.html
I get .8186 between 24 and 48
so
about 819 out of 1000
To determine how many students would finish the test in between 24 and 48 minutes, we need to use the properties of the normal distribution. Specifically, we will calculate the probability that a student's test time falls within this range.
To start, we need to standardize the values 24 and 48 using the formula for standardization:
Z = (X - μ) / σ
where Z is the standardized score, X is the value we want to standardize, μ is the mean of the distribution, and σ is the standard deviation.
For 24 minutes:
Z1 = (24 - 40) / 8 = -2
For 48 minutes:
Z2 = (48 - 40) / 8 = 1
Next, we can use a standard normal distribution table or a calculator with built-in functions to find the probability associated with these standardized values.
Using the standard normal distribution table, we find that the probability associated with Z1 = -2 is approximately 0.0228, and the probability associated with Z2 = 1 is approximately 0.8413.
Now, to find the probability that a student's test time falls between 24 and 48 minutes, we subtract the two probabilities:
P(24 ≤ X ≤ 48) = P(Z1 ≤ Z ≤ Z2) = P(Z ≤ Z2) - P(Z ≤ Z1)
= 0.8413 - 0.0228
≈ 0.8185
Finally, to determine how many students would be expected to finish the test in between 24 and 48 minutes, we multiply the probability by the total number of students:
Expected students = Probability * Total number of students
= 0.8185 * 1,000
≈ 818.5
Since we cannot have a fraction of a student, we can round our answer to the nearest whole number. Therefore, we would expect approximately 819 students to finish the test between 24 and 48 minutes.