You have a score on a national entrance exam of 95. What percentage of people in the population have a score that is lower than yours? (Assume the population mean is 75 and the standard deviation is 10. Use the table I passed out in class.)
Z = (score-mean)/SD
Either use the handout or find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
ewrw 6546
To determine the percentage of people in the population who have a score lower than yours on the national entrance exam, you will need to use the Z-score and the z-table.
The Z-score is a measure that helps to standardize data by expressing it in terms of the number of standard deviations from the mean. It shows how far away a particular observation is from the mean in terms of standard deviations.
First, let's calculate the Z-score. The formula for the Z-score is:
Z = (X - μ) / σ
Where:
X = your score (95 in this case)
μ = population mean (75 in this case)
σ = population standard deviation (10 in this case)
Substituting the given values:
Z = (95 - 75) / 10
Z = 20 / 10
Z = 2
Now, we need to find the corresponding percentage in the z-table. The z-table provides the cumulative probability up to a specific Z-score.
If you refer to the z-table, locate the row that corresponds to 2.0 and look for the column that corresponds to the second decimal place (0.00). The value in the intersection of the row and column is 0.9772.
However, we need to find the percentage of people with scores lower than yours, so we subtract this value from 1 (since the cumulative probability is for all scores up to and including the Z-score).
1 - 0.9772 = 0.0228
Therefore, approximately 2.28% of people in the population would have a score lower than yours on the national entrance exam.