In: Statistics and Probability
If the mean grade on Test #1 was 81.6 with a standard deviation of 4.2, and assuming the distribution of scores was normal, use Table 8.1 to find the 80th percentile.
Solution :
Given that,
mean = = 81.6
standard deviation = = 4.2
Using standard normal table ,
P(Z < z) = 80%
P(Z < 0.84) = 0.8
z = 0.84
Using z-score formula,
x = z * +
x = 0.84 * 4.2 + 81.6 = 85.13
80th percentile is 85.13