The average score for a business examination comprising 300 students is 54%. If the marks are normally distributed with a standard deviation of 8%, determine the number of students scores above 70%.
well, 70 is 2 std above the mean.
Look that up in your Funk and Wagnall's (or, more productively, in your Z table).
Is there a hypothesis you would like to see tested? Comment on why it has not been tested, and briefly describe how you would get the data for the test.
To determine the number of students who scored above 70%, you need to use the properties of the normal distribution.
Step 1: Convert the average score and the score of 70% to z-scores.
The z-score formula is:
z = (x - μ) / σ
Where:
x = score
μ = mean (average)
σ = standard deviation
For the average score:
z_average = (average - μ) / σ
z_average = (54 - 70) / 8
z_average = -16 / 8
z_average = -2
For the score of 70%:
z_70 = (70 - μ) / σ
z_70 = (70 - 70) / 8
z_70 = 0 / 8
z_70 = 0
Step 2: Find the proportion of scores above 70% using the standard normal distribution table or a calculator that can calculate probabilities for a normal distribution.
Using the standard normal distribution table (also known as the Z-table), look up the proportion of scores to the right of z_70. The proportion of scores to the right of z_70 is the same as the proportion of scores above 70%.
The proportion of scores above 70% = 1 - (proportion to the left of z_70)
For z_70 = 0, the proportion to the left is 0.5000 (from the Z-table).
The proportion of scores above 70% = 1 - 0.5000
The proportion of scores above 70% = 0.5000
Step 3: Find the number of students above 70% by multiplying the proportion above 70% by the total number of students.
Number of students above 70% = proportion above 70% * total number of students
Number of students above 70% = 0.5000 * 300
Number of students above 70% = 150
Therefore, the number of students who scored above 70% is 150.