Question from 2018 paper

I have a question from the 2018 paper Q1 b)

So I was just wondering if my explanation would be valid because I did it differently from the marking schedule.

So IQR is the middle 50%; therefore, on a standard normal curve, it would cover an area of 0.25 on either side of the mean, which is 0, so it would cover a total area of 0.5

On a standard normal curve, one standard deviation would cover an area of 0.34 from the mean, which is 0.

0.5> 0.34

0.5/0.34 = IQR is approximately 1.5 times greater than IQR for any normal distribution

Would my working be correct?


Hi @11111111

I can definitely follow your reasoning and think you have made a good argument. I think you were meant to say that the IQR is about 1.5 times greater than 1 SD for any normal distribution.
Unfortunately the schedule states that you can not get higher than Merit by looking at the probability alone. They really wanted you to find the z-values, however I personally feel you have answered the question as it is fairly obvious that 0.5 is greater than 0.34. To secure the higher grade you could have looked up what z-values go with the IQR i.e. find (a<Z<b)=0.5

Often there are different ways of interpreting and answering the question than that of the marking schedule!

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Also in your answer you mixed up percentage of the area under the curve. 0.34 is the area under the curve covered with st. dev = 1 from one side of the mean value (compare to 0.25 from IQR). So you either should have said that 0.34 is greater than 0.25 0.34/0.25=1.36 times or that 0.68 is greater than 0.5 0.68/0.5=1.36 times.

Thanks @Freydis I didn’t read the question properly!

Thank you @Freydis and @y-equals-mx-plus-c for your help!

I am just a little confused about the 0.68 part of your calculations. Because won’t 0.68 be worth two standard deviations when we need to compare 1 SD to the IQR. I have attached the image I have been referring to.


Hi @11111111

Just be careful how you read the question. As it is asking you to “Show that, for any normal distribution, the inter-quartile range (IQR) is always greater than the standard deviation”

So we need to remember that the IQR is the length UQ-LQ and it also holds 50% of the data. You might like to look at the diagram below:

So the length of the IQR when we put it in the standard normal works out to be 1.348 standard deviations. Meaning it will always be larger than the standard deviation for all normal distributions.

I can see how you have got the 34% by thinking of it as just one standard deviation, but the question is asking about “the” standard deviation not specifically one standard deviation.

Hope this makes sense

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