Estimate length using compensation
1 answer:
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The answer is 96%.
Explanation:
It is generally presumed that the scores are normally distributed.
1) You are given how many standard deviations from the mean Jeremy's score is. This is exactly the definition of the z-score. Therefore z = 1.75
2) Look at a left-tail z-table in order to find the area of the normal curve on the left of your z-score (see picture attached). A = 0.9599
3) Multiply the area by 100 in order to find the percentile:
<span>0.9599 </span>× 100 = 95.99
Therefore, 95.99% of the students scored less than Jeremy.
Hence, the answer is 96%.
Explanation:
It is generally presumed that the scores are normally distributed.
1) You are given how many standard deviations from the mean Jeremy's score is. This is exactly the definition of the z-score. Therefore z = 1.75
2) Look at a left-tail z-table in order to find the area of the normal curve on the left of your z-score (see picture attached). A = 0.9599
3) Multiply the area by 100 in order to find the percentile:
<span>0.9599 </span>× 100 = 95.99
Therefore, 95.99% of the students scored less than Jeremy.
Hence, the answer is 96%.