let Ozzie's completion time be 'a'
could you please elaborate on this a bit? I am a bit confused.As the question says he completes ahead of 84% of competitors, it will be 50% (half of the normal distribution) + 34%, meaning that the z-score is 1
Thanks! Can you please elaborate on the z-score part? I thought 68% of scores lie between -1 and 1 so shouldn't it be -3 and 3 to include the 84%? Soz, don't remember anything about this topic.
P(Z<a) = 0.840, that means the percentage of scores below "a" is 0.840could you please elaborate on this a bit? I am a bit confused.
yess thank you for your help i understand it now!P(Z<a) = 0.840, that means the percentage of scores below "a" is 0.840
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The total area under a normal distribution curve is just 1 (or 100%), that means half of it will be 50% (from the left of 0)
Since 68% of scores lie between -1 < z < 1 (according to the empirical rule), the percentage of scores between 0 < z < 1 will be 34%
Now add them together and that produces 50% + 34% which is 84% (ie the percentage of scores Ozzie beats)
Which means the corresponding z-score is 1 so P(Z<1) = 0.840
Okay, in a race if someone finished ahead of a certain amount of competition then they must have finished the race with less time than their competitors. In that case, what should happen is that suppose Ozzie finished ahead of 16% of the competitors then he would have 65 minutes because he is slower than 84% of the competitors. But in this question he has finished ahead of 84% of competitors then what should happen is that he would have finished in 55 minutes because 5 minutes is the standard deviation and using the empirical rule 60-5 makes 55 minutes using the fact that 60 minutes is the mean.