A large number of applicants for admission to graduate study in business are given an aptitude test with a resulting mean score of 460 with a standard deviation of 80. The top 2.5 percent of the applicants would have a score of at what?
A. 606
B. 617
C. 600
D. 646
97.5% must be below
or in your "tables" find the probability closest to
0.975
You should have a z-score of 4.9625 depending on the accuracy of your tables
then (x - 460)/80 = 1.9625
x-460 = 157
x = 157+460 = 617
OR, use this wonderful webpage:
http://davidmlane.com/hyperstat/z_table.html
Click on the second option: value under an area
enter 0.975 for "Area" --- (from 100% - 2.5% = 97.5%)
enter 460 for mean
enter 80 for SD
click on "below" and recalculate to get
616.832 , round off to 617
617
To find the score that corresponds to the top 2.5 percent of applicants, we need to find the value that cuts off the upper tail of the distribution. This value can be found using the Z-score formula.
The Z-score formula is given by:
Z = (X - μ) / σ
Where:
- X is the value we want to find (score in this case)
- μ is the mean of the distribution (460)
- σ is the standard deviation of the distribution (80)
To find the Z-score that corresponds to the top 2.5 percent, we need to find the Z-score that cuts off the lower 97.5 percent of the distribution.
Using a Z-table or calculator, we find that the Z-score corresponding to a cumulative probability of 0.975 is approximately 1.96.
Now, we can rearrange the Z-score formula to solve for X:
X = Z * σ + μ
X = 1.96 * 80 + 460
X ≈ 156.8 + 460
X ≈ 616.8
So, the top 2.5 percent of applicants would have a score of approximately 617.
Therefore, the answer is B. 617.
To find the score that corresponds to the top 2.5 percent of applicants, you will need to use the concept of z-scores.
A z-score tells you how many standard deviations an observation is from the mean. The formula to calculate a z-score is:
z = (x - μ) / σ
Where:
- x is the raw score (in this case, the score you are trying to find)
- μ is the population mean
- σ is the standard deviation of the population
In this case, the mean score (μ) is given as 460, and the standard deviation (σ) is given as 80. We want to find the score that corresponds to the top 2.5 percent, which means we need to find the z-score that corresponds to the 97.5th percentile.
To find the z-score for the 97.5th percentile, we can use a standard normal distribution table or a calculator.
Looking up the z-score in the normal distribution table, you will find that the z-score for the 97.5th percentile is approximately 1.96.
Now, we can use the z-score formula to find the score (x):
1.96 = (x - 460) / 80
Now, solve for x:
1.96 * 80 = x - 460
156.8 = x - 460
x = 156.8 + 460
x ≈ 617.8
So, the score that corresponds to the top 2.5 percent of applicants is approximately 617.8.
Therefore, the answer is B. 617.