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Hi John, I have a little problem with this question.
The results of an accountancy exam are normally distributed with a mean score of 58 and a standard deviation of 10.
What is the percentage probability that a student will score more than 80 (to 2 decimal places)?—>Correct Answer
The percentage probability that a student will score more than 80 is 1.39%
Calculate the Z score:
Z-score = (x – µ)/?
Z = (80 – 58)/10 = 2.2
Look up this number on the standard normal distribution table: 0.4861
This gives us the area under the normal curve between the mean and 80. The area under the curve above 80 will therefore be 0.5 – 0.4861 = 0.0139, or 1.39%
**I understand everything but the part of “the area under the curve above 80 will be 0.5”
How should I calculate that to end with 0.5?
Thank you.Because the normal curve is symmetrical, then 50% (or 0.5) will be above the mean and 50% (or 0.5) will be below the mean.
Therefore 50% will be above the mean of 58.
The answer you quote is not saying that the area under the curve above 80 will be 0.5. What is actually says is that it will be 0.5 (the area above the mean of 58) minus 0.4861 (the area between 58 and 80).
Have you watched my free lectures on this?
Hi John,
After reading your response, I watched again your lecture of “Normal Distribution”, in the minute 20:50 you explained it perfectly… I guess my brain wasn’t at 100% yesterday.
Thank you very much!
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