Scores on an exam are normally distributed with a mean of 76 and a standard deviation of 10. In a group of 230 tests, how many students score above 96?
First, we need to standardize the score of 96 using the formula:
z = (X - μ) / σ
Where z is the standard score, X is the score, μ is the mean, and σ is the standard deviation.
Substituting the given values, we get:
z = (96 - 76) / 10 = 2
Now, we need to find the proportion of scores above 96, which is equivalent to finding the proportion of scores with a z-value greater than 2. We can look up this proportion in the standard normal distribution table.
Looking up the z-value of 2 in the standard normal distribution table, we find that the proportion of scores above 96 is approximately 0.0228.
To find the number of students who score above 96, we multiply this proportion by the total number of tests:
Number of students scoring above 96 = 0.0228 * 230 ≈ 5.24
Therefore, there are approximately 5 students who score above 96.