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50% of the class performs above the mean. 47% of the class, above the mean, does not achieve this level.
From a normal curve table, the z-score that corresponds to 47% from the value to the mean is a z-score of 1.88. Since the z-score represents the number of standard deviations, the value we are looking for is 175 words + (20 * 1.88) = 175 + 37.6 = 212.6.
Your entry into the Final Exam Shell should be 213.
From a normal curve table, the z-score that corresponds to 47% from the value to the mean is a z-score of 1.88. Since the z-score represents the number of standard deviations, the value we are looking for is 175 words + (20 * 1.88) = 175 + 37.6 = 212.6.
Your entry into the Final Exam Shell should be 213.
If Mrs. smith's reading class can read a mean of 175 words per minute with a standard deviation of 20 words per minute. The top 3% of the class is to receive a special award. The minimum number of words per minute a student would have to read in order to get the award will be: 213 words per minute
Using this formula
Minimum number of words =(z score× standard deviation) + means
Where:
z score=(100%-3%=97%)
z score= 97%=1.881
Standard deviation=175
Mean=20
Let plug in the formula
Minimum number of words =(1.881×20) + 175
Minimum number of words =37.62+175
Minimum number of words =212.62
Minimum number of words =213 (Approximately)
Inconclusion if Mrs. smith's reading class can read a mean of 175 words per minute with a standard deviation of 20 words per minute. The top 3% of the class is to receive a special award. The minimum number of words per minute a student would have to read in order to get the award will be: 213 words per minute
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