Refer to the Teaching Psychology (May 1998) study of how external clues influence performance. Two different forms of a midterm psychology examination were given, one printed on blue paper and the other on red paper. Grading only the difficult questions, the researchers found that scores on the blue exam had a distribution with a mean of 53% and a standard deviation of 15%, while scores on the red exam had a distribution with a mean of 39% and a standard deviation of 12%. Assuming that both distributions are approximately normal, on which exam is a student more likely to score below 20% on the difficult questions, the blue one or the red one?

So for this case we have a higher probability for the red exam so we can conclude that the student is more likely to score below 20% on the difficult questions in the red one.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the scores for the blue exam, and for this case we know the distribution for X is given by:

[tex]X \sim N(53,15)[/tex]

Where [tex]\mu=53[/tex] and [tex]\sigma=15[/tex]

We are interested in the probability that P(X<20%)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

And if we replace we got:

[tex] P(X<20) = P(Z<\frac{20-53}{15}) = P(Z<-2.2) = 0.0139[/tex]

Let Y the random variable that represent the scores for the red exam, and for this case we know the distribution for X is given by:

[tex]Y \sim N(39,12)[/tex]

Where [tex]\mu=39[/tex] and [tex]\sigma=12[/tex]

We are interested in the probability that P(Y<20%)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{y-\mu}{\sigma}[/tex]

And if we replace we got:

[tex] P(Y<20) = P(Z<\frac{20-39}{12}) = P(Z<-1.58) = 0.057[/tex]

So for this case we have a higher probability for the red exam so we can conclude that the student is more likely to score below 20% on the difficult questions in the red one.