Can anyone help with number 9?
1 answer:
You might be interested in
Answer:
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:
Where and
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:
And if we replace we got:
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:
Where and
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:
And if we replace we got:
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.