Please answer this as soon as possible.
2 answers:
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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%.
Answer: 14.7338936% decrease
cross-multiply
85.2661064 = x
100 - 85.2661064 = 14.7338936
Check: 0.852661064 * 5355 = 4566
If permitted, you may round for a simpler (yet less precise and exact) answer. For instance, rounding to the hundredths would make it ~14.73%.