A teacher decides to fail 5% of the class. Exam marks are roughly normally distributed
with μ = 72 𝑎𝑛𝑑 σ = 6. What mark must a student make to pass?
You can play around with Z table stuff at
davidmlane.com/hyperstat/z_table.html
Well, to pass the class, the student needs to score better than 5% of the class. Since the exam marks are roughly normally distributed, we can use the concept of z-scores to find the corresponding mark.
To calculate the z-score, we can use the formula: z = (x - μ) / σ, where x is the desired mark, μ is the mean, and σ is the standard deviation.
In this case, we want to find the z-score that corresponds to the lower 5% of the class (since the teacher wants to fail 5%). Let's call this z-score z_pass.
Using a z-table or calculator, we can find that the z-score corresponding to the lower 5% is approximately -1.645.
Now, let's solve for x in the z-score formula:
-1.645 = (x - 72) / 6
Multiplying both sides by 6, we get:
-9.87 = x - 72
Adding 72 to both sides:
x = 72 - 9.87
x ≈ 62.13
So, to pass the class, the student needs to score approximately 62.13 marks or higher.
To find the mark a student must make to pass, we need to calculate the cutoff mark for passing.
Step 1: Convert the given percentage to a z-score.
The teacher wants to fail 5% of the class, which means only the lowest 5% of the scores will be considered as failing. In a normal distribution, we can find the z-score corresponding to a given percentage using a standard normal distribution table or calculator.
For a one-tailed test (since we are only looking at the lowest 5% of scores), 5% is equivalent to a z-score of approximately -1.645.
Step 2: Use the z-score to find the corresponding raw score.
Once we have the z-score, we can find the corresponding raw score using the formula:
Raw score = (z-score * standard deviation) + mean
Given:
Mean (μ) = 72
Standard deviation (σ) = 6
Z-score = -1.645
Using the formula:
Raw score = (-1.645 * 6) + 72
Step 3: Calculate the raw score.
Simplifying the equation, we get:
Raw score = -9.87 + 72
Raw score ≈ 62.13
Therefore, a student must make a mark of approximately 62.13 to pass.
To determine the mark a student must make to pass, we need to find the score below which only 5% of the class falls.
Here's how you can calculate it step-by-step:
Step 1: Standardize the desired percentile using the Z-score formula.
Z = (X - μ) / σ
Where:
Z = Z-score
X = desired score
μ = mean (average)
σ = standard deviation
Step 2: Find the Z-score corresponding to the desired percentile.
We want to find the Z-score for the 5th percentile, which corresponds to a cumulative probability of 0.05. You can look up this value in a standard normal distribution Z-table or use a calculator.
Step 3: Solve the Z-score formula for the desired score.
Rearrange the Z-score formula to solve for X:
X = (Z * σ) + μ
Step 4: Calculate the desired score.
Substitute the value of the Z-score from Step 2 along with the given mean (μ) and standard deviation (σ) into the formula from Step 3. Compute the equation to get the final result.
So, to determine the mark a student must make to pass, follow these steps:
Step 1: Standardize the desired percentile using the Z-score formula:
Z = (X - μ) / σ
X = unknown
μ = 72
σ = 6
Step 2: Find the Z-score for the 5th percentile.
Step 3: Solve the Z-score formula for the desired score:
X = (Z * σ) + μ
Step 4: Calculate the desired score using the values from Step 2:
X = (Z * 6) + 72
By following these steps, you can find the mark a student must make to pass based on the given information.