The combined math and verbal scores for students taking a national standardized examination for college admission, is normally distributed with a mean of 620 and a standard deviation of 220. If a college requires a student to be in the top 10 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?
answer:(round to the nearest integer)
using my favourite normal distribution graphing calculator (same as tables in back of text books)
http://davidmlane.com/hyperstat/z_table.html
click on
"value from an area"
then enter:
area = .9 (to be in top 10%, 90 must be "below" )
mean = 620
SD = 220
click on "below" and recalculate to get 901.98 or 902
Well, if we want to find the minimum score that would place a student in the top 10%, we need to find the score that is higher than 90% of the other scores.
To do this, we can use z-scores and the standard normal distribution table. The z-score is calculated by subtracting the mean from the desired score and dividing it by the standard deviation.
The z-score that corresponds to the top 10% can be found by looking up the cumulative probability of 0.9 in the standard normal distribution table.
Using the formula: z = (X - mean) / standard deviation
where X is the desired score, mean = 620, and standard deviation = 220.
So, z = (X - 620) / 220
Using the standard normal distribution table, we find that the z-score corresponding to the top 10%, or 0.9, is approximately 1.28.
Now we can solve for X:
1.28 = (X - 620) / 220
Multiplying both sides by 220, we get:
281.6 = X - 620
Adding 620 to both sides, we find:
X = 901.6
Rounding to the nearest integer, the minimum score a student would need to qualify for admission at the college is 902.
But hey, don't sweat it! Just remember, standardized tests are just a part of the admissions process. So even if you don't quite hit that score, there are still plenty of other factors that colleges consider. And who knows, maybe you'll be the next famous clown scientist instead!
To find the minimum score that a student must obtain to be in the top 10% of students taking the test, we need to determine the z-score associated with the top 10% of the distribution.
Step 1: Calculate the z-score corresponding to the top 10%:
The top 10% of the distribution corresponds to a percentile of 90%. Using a standard normal distribution table or calculator, we find that the z-score associated with a percentile of 90% is approximately 1.28.
Step 2: Calculate the minimum score:
The z-score formula is given by: z = (x - μ) / σ
Where:
- x is the raw score
- μ is the mean of the distribution (620 in this case)
- σ is the standard deviation of the distribution (220 in this case)
Since we are looking for the minimum score, we need to solve for x in the formula:
1.28 = (x - 620) / 220
Rearranging the formula, we get:
x - 620 = 1.28 * 220
x - 620 = 281.6
x = 620 + 281.6
x ≈ 901.6
Therefore, the minimum score that a student must obtain to qualify for admission at the college is approximately 902 (rounded to the nearest integer).
To find the minimum score that a student can obtain and still qualify for admission in the top 10%, we need to find the score that corresponds to the 90th percentile of the distribution.
Step 1: Convert the percentile to a z-score.
The z-score formula is calculated using the following formula:
z = (x - μ) / σ
where:
z = z-score
x = raw score
μ = population mean
σ = population standard deviation
Since we have the mean (μ = 620) and standard deviation (σ = 220), we can use the z-score formula to calculate the z-score that corresponds to the 90th percentile.
Step 2: Look up the z-score in the standard normal distribution table.
The standard normal distribution table provides the cumulative probability up to a certain z-score. We need to find the z-score that corresponds to the 90th percentile.
By looking up the standard normal distribution table, we can find that the z-score corresponding to the 90th percentile is approximately 1.28.
Step 3: Convert the z-score back to a raw score.
Now that we have the z-score, we can convert it back to a raw score using the formula:
x = (z * σ) + μ
Substituting the values into the formula:
x = (1.28 * 220) + 620
x ≈ 860.6
The minimum score a student can obtain and still qualify for admission at the college is approximately 861 (rounded to the nearest integer).