The result of an exam score for a given class is normally distributed. If the mean score is
85 points and the standard deviation is equal to 20 points, find the cutoff passing grade
such that 83.4% of those taking the test will pass.
you can play around with Z-table stuff here:
http://davidmlane.com/hyperstat/z_table.html
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To find the cutoff passing grade, we need to determine the score value at which 83.4% of the scores fall below that value.
Since the exam score is normally distributed, we can use the standard normal distribution to find the corresponding z-score.
The z-score represents the number of standard deviations a value is from the mean. It allows us to convert the given percentage to a z-score and then use the z-score to find the corresponding score value.
1. Start by finding the z-score using the percentage.
The area under the normal curve to the left of a given z-score represents the percentage of scores below that value.
Using the z-table or a calculator, we can find the z-score that corresponds to 83.4%. For simplicity, we'll use a z-table.
Looking up the z-score for 83.4% in the z-table, we find that the closest value is 0.91.
2. Convert the z-score to the corresponding score value.
The z-score formula is: z = (x - μ) / σ
Where:
- z is the z-score
- x is the score value
- μ is the mean score
- σ is the standard deviation
Rearranging the formula to solve for x:
x = z * σ + μ
Plugging in the values:
x = 0.91 * 20 + 85
x = 18.2 + 85
x = 103.2
Therefore, the cutoff passing grade would be approximately 103.2 points, such that 83.4% of those taking the test will pass.