Question

Urgent!! Need an answer ASAP!

Professor Scott has 84 students in his college mathematics lecture class. The scores on the midterm exam are normally distributed with a mean of 72.3 and a standard deviation of 8.9. How many students in the class can be expected to receive a score between 63.4 and 81.2? Express the answer to the nearest student.

Answer 1

Answer: 57

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Step-by-step explanation:

Convert each raw score to a z score

If x = 63.4, then

z = (x-mu)/sigma

z = (63.4-72.3)/8.9

z = -1

if x = 81.2, then

z = (x-mu)/sigma

z = (81.2-72.3)/8.9

z = 1

So P(63.4 < x < 81.2) is the same as P(-1 < z < 1)

Through the empirical rule (aka the 68-95-99.7 rule), we know that roughly 68% of the data values are within one standard deviation of the mean. In other words

P(-1 < z < 1) = 0.68 approximately (see note below)

Multiply 0.68 by the total 84 to get

0.68*84 = 57.12

which rounds to 57 people

Note: we can use a calculator to get a more accurate result of 0.68268949; a data table works as well but it won't be as accurate. For the purposes of this problem, the value 0.68 is good enough

Answer 2

Answer: 57

=====================================================

Step-by-step explanation:

Convert each raw score to a z score
If x = 63.4, then
z = (x-mu)/sigma
z = (63.4-72.3)/8.9
z = -1

if x = 81.2, then
z = (x-mu)/sigma
z = (81.2-72.3)/8.9
z = 1

So P(63.4 < x < 81.2) is the same as P(-1 < z < 1)

Through the empirical rule (aka the 68-95-99.7 rule), we know that roughly 68% of the data values are within one standard deviation of the mean. In other words
P(-1 < z < 1) = 0.68 approximately (see note below)

Multiply 0.68 by the total 84 to get
0.68*84 = 57.12
which rounds to 57 people

Note: we can use a calculator to get a more accurate result of 0.68268949; a data table works as well but it won't be as accurate. For the purposes of this problem, the value 0.68 is good enough