A sociology professor assigns letter grades on a test according to the following scheme.
A: Top 14% of scores
B: scores below the top 14% and above the bottom 55%
C: scores below the top 45% and above the bottom 17%
D: scores below the top 83% and above the bottom 6%
F: bottom 6 % of scores
Scores on the test are normally distributed with a mean of 78.3 and a standuard deviation of 7.3. Find the numerical limits for a B grade. Round your answers to the nearest whole number if necessary.
Can someone please help and show all work thank you so much!
To find the numerical limits for a B grade, we need to first determine the corresponding Z-scores.
Step 1: Find the Z-score for the top 14%.
To find the Z-score, we can use the formula:
Z = (X - μ) / σ
where:
X = the score we want to find the Z-score for (top 14%)
μ = the mean (78.3)
σ = the standard deviation (7.3)
To find the score corresponding to the top 14%, we need to find the Z-score that has an area of 0.86 to the left on the standard normal distribution table. Looking up the Z-score in the table, we find it to be approximately 1.08.
Step 2: Find the Z-score for the bottom 55%.
Similarly, we need to find the Z-score that has an area of 0.45 to the left on the standard normal distribution table. Looking up the Z-score, we find it to be approximately -0.13.
Step 3: Convert the Z-scores back to scores.
Using the Z-score formula, we can rearrange it to solve for X:
X = Z * σ + μ
For the top 14%:
X = 1.08 * 7.3 + 78.3
X ≈ 86.41
For the bottom 55%:
X = -0.13 * 7.3 + 78.3
X ≈ 77.75
Therefore, the numerical limits for a B grade are approximately 78 (rounded down) and 86 (rounded up).
To find the numerical limits for a B grade, we need to determine the scores that fall below the top 14% and above the bottom 55% of all scores.
Step 1: Convert the given information into z-scores.
To do this, we will use the formula:
z = (x - μ) / σ
where x is the given score, μ is the mean, and σ is the standard deviation.
Step 2: Find the z-score corresponding to the top 14% of scores.
We want to find the z-score that corresponds to the cumulative area of 14% from the mean to the right tail of the distribution. We can use a standard normal distribution table (also known as a z-table) or utilize statistical software for this calculation.
Using the z-table, we find that the z-score for a cumulative area of 0.14 is approximately 1.08.
Step 3: Find the z-score corresponding to the bottom 55% of scores.
We want to find the z-score that corresponds to the cumulative area of 55% from the left tail of the distribution to the mean. This can also be calculated using a z-table or statistical software.
Using the z-table, we find that the z-score for a cumulative area of 0.55 is approximately -0.11.
Step 4: Find the corresponding raw scores.
To find the corresponding raw scores, we will use the formula:
x = μ + (z * σ)
where x is the raw score, μ is the mean, z is the z-score, and σ is the standard deviation.
For the upper limit of the B grade:
x_upper = μ + (1.08 * 7.3)
= 78.3 + 7.884 (rounded to the nearest whole number)
≈ 86 (rounded to the nearest whole number)
For the lower limit of the B grade:
x_lower = μ + (-0.11 * 7.3)
= 78.3 - 0.803 (rounded to the nearest whole number)
≈ 77 (rounded to the nearest whole number)
Therefore, the numerical limits for a B grade are approximately 77 to 86.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities and their Z scores.
Z = (score-mean)/SD