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3000 students take a college entrance exam. The scores on the exam have an approximately normal...

3000 students take a college entrance exam. The scores on the exam have an approximately normal distribution with mean mu equals53 points and standard deviation sigma equals10 points. Use the? 68-95-99.7 rule to complete the following. a. Estimate the percentage of students scoring 53 points or less . b. Estimate the percentage of students scoring 73 points or more .

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Answer #1

Given, mean = 53, standard deviation = 10 and number of students = 3000.

By 68-95-99.7 rule we know that :

68% of data is within 1 standard deviation () of the mean ()

95% of data is within 2 standard deviation () of the mean ()

99.7% of data is within 2 standard deviation () of the mean ()

a) Here we have to find the percentage of students scoring 53 points or less.

i.e., the percentage of data within 0 and the mean ().

Therefore, the required percentage (from the graph) is = (0.13+2.14+13.59+34.13)% 50%.

b) Here we have to find the percentage of students scoring 73 points or more.

The interval is .

Therefore, the required percentage (from the graph) is = (2.14+0.13)% 2.3%.

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