If only the upper 30% of a normally distributed class passed a quiz for which the mean was 70 and the standard deviation was 10, what was the lowest score a student could have received and still have passed?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.30) related to a Z score. Use the equation above to find the score.
64.8
To find the lowest score a student could have received and still have passed, we need to determine the z-score associated with being in the upper 30% of the distribution.
Step 1: Convert the desired percentile to a z-score.
Since we want the upper 30% of the class, we subtract 30% from 100% to find the lower 70% of the distribution. So, the desired percentile corresponds to the z-score that encloses the lower 70% of the distribution.
Step 2: Use the z-score to find the corresponding x-value using the mean and standard deviation.
The formula to convert a z-score to an x-value is:
x = (z * standard deviation) + mean
Given:
Mean (μ) = 70
Standard Deviation (σ) = 10
Step 1: Find the z-score using the standard normal distribution table or calculator.
Since we want to find the z-score for the lower 70%, we look up the z-score associated with the upper 30%. Using a standard normal distribution table, the z-score is approximately 0.524.
Step 2: Calculate the lowest score a student could have received.
x = (z * standard deviation) + mean
x = (0.524 * 10) + 70
x ≈ 5.24 + 70
x ≈ 75.24
Therefore, the lowest score a student could have received and still have passed is approximately 75.24.
To find the lowest score a student could have received and still have passed, we need to determine the cutoff score that separates the upper 30% from the rest of the class.
First, let's calculate the z-score corresponding to the upper 30% of the distribution. The z-score measures the number of standard deviations a particular value is away from the mean.
Using a standard normal distribution table or a calculator, we can find that the z-score corresponding to the upper 30% is approximately 0.524 (rounded to three decimal places).
Next, we can use the formula for z-score to determine the actual score corresponding to this z-score:
z = (x - mean) / standard deviation
Rearranging the formula and plugging in the known values:
0.524 = (x - 70) / 10
Now, solve for x (the lowest score a student could have received):
0.524 * 10 = x - 70
5.24 = x - 70
x = 75.24
Therefore, the lowest score a student could have received and still have passed the quiz is approximately 75.24 (rounded to two decimal places).