the marks of 100 students in an examination are normally distributed with mean of 25 marks and a standard deviation of 10 marks. Given that the pass mark is 20, estimate the number of students who passed the examination.
The normal distribution is N(25, 10).
P(X > 20) = .691
So approximately 69.1% passed the exam.
69.1% of 100 students is 69 students.
To estimate the number of students who passed the examination, we need to find the proportion of students whose marks are above the pass mark of 20.
In a normal distribution, we can use the z-score formula to determine the proportion of values above a given cutoff point. The z-score formula is:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
Let's calculate the z-score for the pass mark:
z = (20 - 25) / 10 = -0.5
Now, we need to find the proportion of values above this z-score. We can use a normal distribution table or a statistical calculator to find this proportion.
Using a normal distribution table, we can find the proportion corresponding to a z-score of -0.5. In many tables, the value for -0.5 is approximately 0.3085.
Since the distribution is symmetrical, we can estimate that the proportion of values below a z-score of -0.5 is also approximately 0.3085.
To find the proportion of students who passed the examination, we subtract the proportion of values below the pass mark from 1:
proportion of students who passed = 1 - 0.3085 = 0.6915
Finally, to estimate the number of students who passed, we multiply the proportion by the total number of students:
number of students who passed = 0.6915 * 100 = 69.15
Since we can't have a fraction of a student, we round the result to the nearest whole number:
number of students who passed ≈ 69
Therefore, an estimated 69 students passed the examination.