scores on a university examination are normally distributed with a mean of 68 and standard deviation of 9. What percetnage of students scores above 77?
The answer is 16% according to the professor that is correct. But I don't understand how the professor got to that answer.
To determine the percentage of students who scored above 77 on the university examination, we can utilize the properties of a normal distribution and calculate the area under the curve.
Step 1: Standardize the score
First, we need to standardize the score of 77 using the formula:
z = (x - μ) / σ
where:
- x is the value we want to standardize (in this case, 77)
- μ is the mean of the distribution (68)
- σ is the standard deviation of the distribution (9)
Substituting the given values:
z = (77 - 68) / 9
z = 9 / 9
z = 1
Step 2: Find the area under the curve
Next, we need to find the area under the normal distribution curve to the right of z = 1, representing scores above 77. This can be done using a z-table or a calculator.
According to standard normal distribution tables or calculators, the area to the right of z = 1 is approximately 0.1587.
Step 3: Convert the area to a percentage
Lastly, we convert the area to a percentage by multiplying it by 100:
0.1587 * 100 = 15.87%
Therefore, approximately 15.87% of students scored above 77 on the university examination. Rounded to the nearest whole number, it is 16% as mentioned by your professor.