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Scores on a certain nationwide college entrance examination follow a normal distribution with a mean of 500 and a standard deviation of 100. Find the probability that a student will score

a) over 650
b) less than 459
c) between 325 and 675
d) If a school only admits students who score over 680, what proportion of the student's pool would be eligible for admission
e) what limit (score) would you set that makes the top 20% of the students eligible

User Porkbutts
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Final answer:

The probability of scoring over 650 is 6.68%, the probability of scoring less than 459 is 33.94%, the probability of scoring between 325 and 675 is 81.86%, the proportion of the student's pool that would be eligible for admission with a score over 680 is 3.59%, and the score that makes the top 20% of the students eligible is 584.

Step-by-step explanation:

a) To find the probability that a student will score over 650, we need to calculate the z-score and use the z-table. The z-score is calculated as (x - mean) / standard deviation. So, the z-score for a score of 650 is (650 - 500) / 100 = 1.5. Using the z-table, we find that the probability of getting a score above 1.5 standard deviations is approximately 0.0668 or 6.68%.

b) To find the probability that a student will score less than 459, we calculate the z-score as (459 - 500) / 100 = -0.41. Using the z-table, we find that the probability of getting a score below -0.41 standard deviations is approximately 0.3394 or 33.94%.

c) To find the probability that a student will score between 325 and 675, we calculate the z-scores for 325 and 675. The z-score for 325 is (325 - 500) / 100 = -1.75, and the z-score for 675 is (675 - 500) / 100 = 1.75. Using the z-table, we find that the probability of getting a score between -1.75 and 1.75 standard deviations is approximately 0.8186 or 81.86%.

d) If a school only admits students who score over 680, we calculate the z-score for 680 as (680 - 500) / 100 = 1.8. Using the z-table, we find that the proportion of the student's pool that would be eligible for admission is approximately 0.0359 or 3.59%.

e) To find the score that makes the top 20% of the students eligible, we need to find the z-score corresponding to the 80th percentile. Using the z-table, we find that the z-score for the 80th percentile is approximately 0.84. We can calculate the score using the formula score = (z-score * standard deviation) + mean. So, the score that makes the top 20% of the students eligible is (0.84 * 100) + 500 = 584.

User Molbdnilo
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