An economics exam is worth 60 points and is taken by 500 students. the scores on the exam are normally distributed with the average of 38 points and a standard deviation of 5 points. Evert, who got 43 out of 60, claims he's in the top 100, is he right?
You can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
To determine if Evert is in the top 100 students based on his score of 43 out of 60, we need to calculate his percentile rank.
First, let's calculate the z-score of Evert's score using the formula:
z = (x - μ) / σ
where x is Evert's score, μ is the average score, and σ is the standard deviation.
In this case, x = 43, μ = 38, and σ = 5.
Using these values:
z = (43 - 38) / 5
z = 5 / 5
z = 1
Next, we need to find the percentile associated with this z-score. We can use a z-table or a statistical calculator to find the corresponding percentile.
A z-score of 1 corresponds to a percentile of approximately 84.1%. This means that Evert scored better than approximately 84.1% of the students who took the exam.
Since there are 500 students who took the exam, the top 100 students would be in the 80th percentile or above.
Therefore, Evert is indeed in the top 100 students as his percentile rank is higher than the 80th percentile.
To determine if Evert is in the top 100 of the economics exam, we need to find out what score corresponds to the 100th rank. We can achieve this by calculating the z-score for Evert's score and comparing it to the z-score for the 100th rank.
Here's how to do it step by step:
1. Find the z-score for Evert's score of 43. The z-score measures how many standard deviations Evert's score is from the mean. The formula for calculating the z-score is:
z = (X - μ) / σ
Where:
X = Evert's score
μ = mean of the distribution
σ = standard deviation of the distribution
Plugging in the values:
z = (43 - 38) / 5
z ≈ 1
2. Look up the z-score in the standard normal distribution table (also called z-table) to find the proportion of the distribution that lies below that z-score. In this case, we are interested in the proportion above that z-score, as we want to know if Evert is in the top 100.
From the z-table, a z-score of 1 corresponds to a proportion of approximately 0.8413.
3. Subtract the proportion obtained in step 2 from 1 to find the proportion above the z-score. In this case, it would be:
P(above z) = 1 - 0.8413
≈ 0.1587
4. Multiply the proportion above the z-score by the total number of students (500) to find the number of students above the z-score.
Number of students above z = Proportion above z * Total number of students
= 0.1587 * 500
≈ 79.35
Since the number of students above Evert's z-score (79.35) is less than 100, Evert's claim of being in the top 100 is incorrect.