The result of an exam score for a given class is normally distributed. If the mean score is 85 points and the standard devotion is equal to 20points find the cutoff passing grade such that 83. 4% of those taking the test will pass
65.6
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To find the cutoff passing grade, we can use the concept of the standard normal distribution, also known as the Z-score.
Step 1: Convert the given percentage to a Z-score
First, we need to find the Z-score corresponding to the percentage of students passing the test. The percentage given is 83.4%. To find the Z-score, we can use a standard normal distribution table or a statistical calculator.
A Z-score represents the number of standard deviations a data point is away from the mean. We are given that the mean is 85 and the standard deviation is 20. We want to find the Z-score such that 83.4% of the data is below it.
Assuming a symmetrical normal distribution, we can find the Z-score by subtracting 83.4% from 100% (to find the area above the Z-score) and dividing by 2 (to find the area from the Z-score to the mean). Let's calculate it:
(100% - 83.4%) / 2 = 8.3%
This means that 8.3% of the scores fall above the Z-score we need to find.
Step 2: Find the Z-score using the standard normal distribution table or calculator
Using a standard normal distribution table or calculator, we can find the Z-score corresponding to the area of 8.3% above it. Let's assume the Z-score is denoted as Z_passing.
From the standard normal distribution table or calculator, we find that the Z-score corresponding to 8.3% above it is approximately 1.09.
Step 3: Convert the Z-score back to an exam score
Now that we have the Z-score, we can find the corresponding exam score. To convert the Z-score back to an exam score, we can use the formula:
X = Z * σ + μ
where X is the exam score, Z is the Z-score, σ is the standard deviation, and μ is the mean.
In this case, we have:
X = 1.09 * 20 + 85
X ≈ 107.8
So, the cutoff passing grade is approximately 107.8 points.