Why a normal distribution is not a good approximation for these exam scores?

[songoku]

Homework Statement

The average score for an exam is 72.5 out of 100 and the standard deviation is 15.3
Why a normal distribution would not give a good approximation to the distribution of the score?

Relevant Equations

Not sure

I am not really sure what the reason is but my argument would be if normal distribution is appropriate, then almost all the score will fall in the range of μ - 3σ to μ + 3σ

For this case, the range of μ - 3σ to μ + 3σ is 26.6 to 118.4 and all the score is unlikely to be within the range.

I feel my argument is not strong enough to justify why normal distribution is not appropriate. Since the max score is 100, the range can be truncated to 26.6 to 100. One can argue that this is a plausible range for exam score

Is there a certain condition to justify use of normal distribution, maybe something like the standard deviation should be at most a % of the mean?

Thanks

 

[FactChecker]

I think you are on the right track but can make a stronger argument. Specifically, what is wrong with the range of 26.6 to 118.4? Concentrate on the problem with that and give a specific probability for the problem values of a normal distribution.

 

[songoku]

I think you are on the right track but can make a stronger argument. Specifically, what is wrong with the range of 26.6 to 118.4? Concentrate on the problem with that and give a specific probability for the problem values of a normal distribution.

For normal distribution to be appropriate:
P(μ - σ < X < μ + σ) = 68.26%

P(μ - 2σ < X < μ + 2σ) = 95.44%

P(μ - 3σ < X < μ + 3σ) = 99.73%In this question:
P(μ - σ < X < μ + σ) =P(57.2 < X < 87.8) = 68.26%

P(μ - 2σ < X < μ + 2σ) = P(41.9 < X < 103.1) = P(41.9 < X < 100) = 94.11%

P(μ - 3σ < X < μ + 3σ) = P(26.6 < X < 118.4) = P(26.6 < X < 100) = 96.25%

Since P(μ - 3σ < X < μ + 3σ) does not match with theoretical value, so normal distribution is not good approximation?

Thanks

 

[FactChecker]

Assuming a normal distribution with that mean and standard deviation, I would calculate the probability, P(X>100), which all are invalid results. It may be a matter of preference, but I think that is the strongest argument.

 

[songoku]

Assuming a normal distribution with that mean and standard deviation, I would calculate the probability, P(X>100), which all are invalid results. It may be a matter of preference, but I think that is the strongest argument.

P(X > 100) = 0.0361= 3.61%

There will be around 3.61% of invalid data in the region of upper tail of normal distribution so normal distribution is not good approximation.

Is this what you mean?

Thanks

 

[FactChecker]

Yes.

 

[songoku]

Thank you very much FactChecker