A math professor notices that scores from a recent exam are normally distributed with a mean of 64 and a standard deviation of 5. Answer the following questions using integer values.
What score do 75% of the students exam scores fall below?
And, how was this answer calculated?
Use the same process as indicated in your later post.
To find the score below which 75% of the students' exam scores fall, we can use the concept of the standard normal distribution.
First, we need to find the z-score associated with a cumulative probability of 75%.
Using a standard normal distribution table or a calculator, we find that a cumulative probability of 75% corresponds to a z-score of approximately 0.674.
Next, we can use the formula for z-score conversion to find the raw score:
z = (x - μ) / σ
where:
z is the z-score,
x is the raw score,
μ is the mean, and
σ is the standard deviation.
Rearranging the formula, we have:
x = z * σ + μ
Plugging in the known values, we get:
x = 0.674 * 5 + 64
Calculating this, we find:
x ≈ 67.37
Since we are looking for an integer score, we round down the calculated value:
x = 67
Therefore, 75% of the students' exam scores fall below 67.
To find the score below which 75% of students' exam scores fall, we need to find the corresponding z-score using the standard normal distribution table or a calculator.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
where:
x = score
μ = mean
σ = standard deviation
In this case, we want to find the z-score that corresponds to the 75th percentile, which is z = 0.674.
Next, we rearrange the z-score formula to solve for the score, x:
x = μ + z * σ
Now, we can plug in the values:
x = 64 + 0.674 * 5
Calculating this, we find:
x ≈ 67.37
Rounding to the nearest integer, approximately 67, the score below which 75% of the students' exam scores fall is 67.
Therefore, 67 is the answer, and it was calculated by finding the z-score corresponding to the 75th percentile and then converting the z-score back to the original score using the mean and standard deviation of the distribution.