a standardized test has a mean of 88 and a standard deviation of 12. What is the score at thr 90th percentile? Assume a normal distribution.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion.
Z = (score-mean)/SD
Plug in the values and solve.
To find the score at the 90th percentile, we need to use the concept of z-scores and the standard normal distribution table.
Step 1: Calculate the z-score using the standard deviation and the mean.
The formula for calculating the z-score is (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. In this case, the mean (μ) is 88 and the standard deviation (σ) is 12.
Step 2: Look up the z-score in the standard normal distribution table.
The standard normal distribution table provides the area under the curve for different z-scores. It tells us what percentage of the data falls below a certain z-score.
To find the z-score that corresponds to the 90th percentile, we need to find the value of z that gives us an area of 0.90 to the left of it in the standard normal distribution table.
Step 3: Calculate the score using the z-score and the formula.
Finally, we can use the formula X = μ + (z * σ) to find the score with the corresponding z-score.
Let's calculate it step by step:
Step 1:
z = (X - μ) / σ
z = (X - 88) / 12
Step 2:
We need to find the z-score that corresponds to the 90th percentile. From the standard normal distribution table, the z-score that gives us an area of 0.90 to the left of it is approximately 1.28.
Step 3:
X = μ + (z * σ)
X = 88 + (1.28 * 12)
X = 88 + 15.36
X ≈ 103.36
Therefore, the score at the 90th percentile is approximately 103.36.