In a Statistics test, the average mark is 82 and the standard deviation is 5. All the students from 88 marks up to 94 marks will receive a B grade. If the marks have a normal distribution and 8 students receive B grade, determine the total number of students who sat for the test.
Question
In a Statistics test, the average mark is 82 and the standard deviation is 5. All the students from 88 marks up to 94 marks will receive a B grade. If the marks have a normal distribution and 8 students receive B grade, determine the total number of students who sat for the test.
Solution
To solve this problem, we need to use the properties of the normal distribution.
Step 1: Calculate the Z-scores for the B grade range (88 to 94 marks). The Z-score is calculated as (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.
For 88 marks: Z1 = (88 - 82) / 5 = 1.2
For 94 marks: Z2 = (94 - 82) / 5 = 2.4
Step 2: Look up these Z-scores in a standard normal distribution table or use a calculator with a normal distribution function to find the proportion of students who scored within this range.
The proportion for Z1 (1.2) is approximately 0.3849 and for Z2 (2.4) is approximately 0.4918.
Step 3: Subtract the proportion of students who scored below 88 from the proportion of students who scored below 94 to find the proportion of students who received a B grade.
P(B) = P(Z2) - P(Z1) = 0.4918 - 0.3849 = 0.1069
Step 4: If 8 students received a B grade, then 0.1069 of the total number of students is 8. Solve for the total number of students.
Total students = 8 / 0.1069 ≈ 75
So, approximately 75 students sat for the test.
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