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.
P(X > 20) = .691
So approximately 69.1% passed the exam.
69.1% of 100 students is 69 students.
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.