a teacher decides to fail 25% of the class.Examination marks are roughly normally distributed w/a mean of 72 & standard deviation of 6. what mark must a student make to pass?
To find out what mark a student must make to pass, we need to determine the cutoff point for passing based on the given information.
Since examination marks are approximately normally distributed with a mean (μ) of 72 and a standard deviation (σ) of 6, we can use the z-score formula to find the cutoff point.
A z-score measures how many standard deviations a given value is from the mean. We can use z-score tables or a calculator to find the cutoff point.
Step 1: Calculate the z-score for the desired percentile.
The teacher wants to fail 25% of the class, which means the passing score will be the score that corresponds to the 75th percentile. The z-score for the 75th percentile is 0.674 (you can look it up in a z-score table or use a calculator).
Step 2: Use the z-score formula to find the raw score for the desired z-score.
The z-score formula is: z = (x - μ) / σ
Rearranging the formula, we have: x = z * σ + μ
Substituting the values, we get: x = 0.674 * 6 + 72
Calculating the equation, x ≈ 75.04
Therefore, a student must make approximately 75.04 marks to pass the class.
To find out the mark a student must achieve to pass, we need to determine the cutoff point that separates the passing students from the failing students.
Since the examination marks are approximately normally distributed with a mean of 72 and a standard deviation of 6, we can use the standard normal distribution table or a statistical calculator to find the cutoff point.
First, we need to convert the 25% failure rate to a z-score. The z-score represents the number of standard deviations a particular value is away from the mean of the distribution.
To find the z-score, we use the formula:
z = (x - μ) / σ
Where:
z = z-score
x = value we want to convert to a z-score (passing mark)
μ = mean of the distribution (72 in this case)
σ = standard deviation of the distribution (6 in this case)
Rearranging the formula to solve for x:
x = z * σ + μ
Now, let's calculate the z-score for the 25% failure rate.
Since the teacher wants to fail 25% of the class, we need to find the z-score that corresponds to the cumulative probability of 0.25 in the standard normal distribution.
Using the standard normal distribution table, we find that the z-score corresponding to a cumulative probability of 0.25 is approximately -0.674.
Now we can calculate the passing mark:
x = -0.674 * 6 + 72
x ≈ 67.35
Therefore, a student must achieve a mark of approximately 67.35 (rounded to the nearest whole number) to pass the class.