SOLUTION: The time required to complete a college achievement test was foundto be normally
distributed, with a mean of 110 minutes and standard deviation of20 minutes.
(a) What proport
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Question 1203585: The time required to complete a college achievement test was foundto be normally
distributed, with a mean of 110 minutes and standard deviation of20 minutes.
(a) What proportion of the students will ?nish within 2 hours (120minutes)?
(b) What proportion of the students will ?nish at exactly 2hours?
(c) When should the test be terminated to allow just enough timefor 90% of the
students to complete the test
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean is 110 minutes.
standard deviation is 20 minutes.
use the z-score formula.
it is z = (x - m) / s
z is the z-score
m is the population mean
s is the standard deviation
x is the student score
The time required to complete a college achievement test was found to be normally distributed, with a mean of 110 minutes and standard deviation of20 minutes.
(a) What proportion of the students will finish within 2 hours (120 minutes)?
z = (120 - 110) / 20 = .5
area to the left of a z-score of .5 = .6915.
this means that 69.15% of the participants will complete the test in 120 minutes or less.
(b) What proportion of the students will finish at exactly 2 hours?
the theory says that, because this is a continuous curve, the probability of getting exactly 120 minutes is 0.
here's refference that addresses that issue.
https://sphweb.bumc.bu.edu/otlt/mph-modules/bs/bs704_probability/bs704_probability8.html#:~:text=Note%20that%20with%20the%20normal,is%20%3C%2029%20is%2050%25.
the particular statement in that reference is:
Note that with the normal distribution the probability of having any exact value is 0 because there is no area at an exact BMI value, so in this case, the probability that his BMI = 29 is 0, but the probability that his BMI is <29 or the probability that his BMI is < 29 is 50%.
(c) When should the test be terminated to allow just enough time for 90% of the
students to complete the test.
z-score that has 90% of the area under the normal distribution curve to the left of it is equal to 1.2816.
z-score formula becomes:
1.2816 = (x - 110) / 20
solve for x to get x = 20 * 1.2816 + 110 = 135.632.
that's the number of minutes that the test would be terminated so that 90% of the participants will have been able to complete the test.
any rounding should be rounding up to ensure at least 90%.
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