3 + 2/9 as a improper fraction and as a decimal
2 answers:
You might be interested in
Answer:
12.92%
Step-by-step explanation:
Mean of the scores= u = 500
Standard deviation = = 10.6
We have to find what proportion of students scored more than 512 marks.
The distribution of scores in a test generally follows the Normal distribution. So we can assume that the distribution of MCAT scores is normally distributed about the mean.
Since, the distribution is normal, we can use the concept of z scores to find the proportion of students who scored above 512.
The formula for z scores is:
So, z score for x = 512 will be:
Thus,
P(X > 512) is equivalent to P(z > 1.13)
So, the test scores of 512 is equivalent to a z score of 1.13. Using the z table we have to find the proportion of z scores being greater than 1.13, which comes out to be 0.1292
Since,
P(X > 512) = P(z > 1.13)
We can conclude that, the proportion of students taking the MCAT who had a score over 512 is 0.1292 or 12.92%