1-the graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard diviation of 100, what would be the cutoff score for selecting the top 10% of applicants, assuming that test is normally distributed?
Convert x=score to z, the standard variable for a normally distributed experiment.
z=(x-μ)/σ
=(x-500)/100
From the normal distribution table,
http://www.math.unb.ca/~knight/utility/NormTble.htm
the z value for 90% cut-off is 1.282, so we have
1.282=(x-500)/100
x=100*1.282+500
=628
To find the cutoff score for selecting the top 10% of applicants, we need to determine the z-score corresponding to the 90th percentile.
The z-score formula is given by:
z = (x - μ) / σ
Where:
- x is the raw score on the test
- μ is the mean of the test scores
- σ is the standard deviation of the test scores
In this case, the mean (μ) is 500 and the standard deviation (σ) is 100.
To find the z-score corresponding to the 90th percentile, we need to find the value of z for which the area under the normal distribution curve to the left of that z-score is equal to 0.90.
Using a z-table or a statistical calculator, we can find that the z-score corresponding to the 90th percentile is approximately 1.282.
Now we can use the z-score formula to find the cutoff score (x) for selecting the top 10% of applicants.
1.282 = (x - 500) / 100
Now, let's solve for x.
Multiply both sides of the equation by 100:
128.2 = x - 500
Add 500 to both sides of the equation:
x = 628.2
Therefore, the cutoff score for selecting the top 10% of applicants is approximately 628.2 on the standardized test.