The distribution of scores on a standardized aptitude test is approximately normal with a mean of 490 and a standard deviation of 95. What is the minimum score needed to be in the top 25% on this test? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest integer.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.25) in smaller portion to get the Z score. Insert values in the equation above.
To find the minimum score needed to be in the top 25% on the test, we need to determine the cutoff score at which 25% of the scores are below and 75% of the scores are above.
Since the distribution is approximately normal, we can use the z-score formula to standardize the scores.
The formula for the z-score is:
z = (x - μ) / σ
Where:
x = score
μ = mean
σ = standard deviation
In this case, the mean (μ) is 490 and the standard deviation (σ) is 95.
To find the z-score corresponding to the top 25% of the distribution, we need to find the z-score that corresponds to a cumulative area of 0.75 (since 75% of the scores will be below).
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative area of 0.75 is approximately 0.6745.
Now, we can use the z-score formula to find the minimum score (x) needed to be in the top 25%:
0.6745 = (x - 490) / 95
Solving for x:
x - 490 = 0.6745 * 95
x - 490 = 64.18175
x = 64.18175 + 490
x = 554.18175
Rounding this to the nearest integer, the minimum score needed to be in the top 25% on the test is 554.
Therefore, to be in the top 25% on this standardized aptitude test, a score of at least 554 is needed.
To find the minimum score needed to be in the top 25%, we need to find the score that corresponds to the 75th percentile.
Step 1: Calculate the z-score corresponding to the 75th percentile.
The z-score is calculated using the formula:
z = (x - μ) / σ
where:
x = score (unknown)
μ = mean = 490
σ = standard deviation = 95
To calculate the z-score corresponding to the 75th percentile, we use the z-table (or a calculator) to find the z-value that corresponds to the cumulative probability of 0.75.
From the z-table, the z-value that corresponds to a cumulative probability of 0.75 is approximately 0.674.
Step 2: Use the z-score formula to find the score corresponding to the z-value.
Rearranging the z-score formula, we have:
x = z * σ + μ
Substituting the values, we have:
x = 0.674 * 95 + 490
Calculating the value, we get:
x = 64.15 + 490
x = 554.15
Step 3: Round the answer to the nearest whole number.
Rounding 554.15 to the nearest whole number, we get:
x ≈ 554
So, the minimum score needed to be in the top 25% on this test is approximately 554.