If GMAT scores for applicants at Oxnard Graduate School of Business are N(500, 50) then the top 5 percent of the applicants would have a score of at least (choose the nearest integer):
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) related to a Z score. Insert the values into the equation below and solve for the score.
Z = (score-mean)/SD
To determine the score at which the top 5 percent of applicants would have a score of at least, we need to find the value at the 95th percentile of the GMAT scores distribution. The given distribution specifies that GMAT scores at Oxnard Graduate School of Business follow a normal distribution with a mean (μ) of 500 and a standard deviation (σ) of 50.
To solve this, we can use a Z-table or calculator to find the Z-score corresponding to the 95th percentile. The Z-score represents the number of standard deviations a value is from the mean.
Using the Z-table, we can find that the Z-score corresponding to the 95th percentile is approximately 1.645. This means that the top 5 percent of applicants will have GMAT scores that are at least 1.645 standard deviations above the mean.
Next, we can calculate the actual score using the formula:
X = μ + (Z * σ)
X = 500 + (1.645 * 50)
X ≈ 582.25
Thus, the top 5 percent of applicants would have a GMAT score of at least 582 (rounded to the nearest integer).
To find the score which corresponds to the top 5 percent of the applicants at Oxnard Graduate School of Business, we can use the Z-score formula.
The Z-score formula is given by:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the value of interest (in this case, the score we want to find)
μ is the mean of the distribution (500 in this case)
σ is the standard deviation of the distribution (50 in this case)
To find the Z-score corresponding to the top 5 percent, we need to find the Z-score that corresponds to a cumulative probability of 0.95 (1 - 0.05).
Using a standard normal distribution table or a calculator, we can find that the Z-score corresponding to a cumulative probability of 0.95 is approximately 1.645.
Now, we can rearrange the Z-score formula to solve for X:
Z = (X - μ) / σ
1.645 = (X - 500) / 50
Multiplying both sides of the equation by 50, we get:
82.25 = X - 500
To solve for X, we add 500 to both sides of the equation:
X = 82.25 + 500 ≈ 582.25
Rounding to the nearest integer, the top 5 percent of the applicants would have a score of at least 582.