If the mean grade on Test #1 was 81.6 with a standard deviation of 4.2, and...

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.

Expert Solution

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

orchestra answered 2 years ago

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