Question

Suppose a professor for biostatistics keeps a record of the final grades reported to the academic institution each semester. Assume the distribution of all final grades varies according to a normal distribution with mean 81.2% and standard deviation 6.4%.

A) What score corresponds to the highest 1% of the distribution?

B) What score corresponds to the lowest 1% of the distribution?

Answer 1

Solution:-

Given that,

mean = = 81.2% = 0.812

standard deviation = = 6.2% = 0.062

a) Using standard normal table,

P(Z > z) = 1 %

= 1 - P(Z < z) = 0.01

P(Z < z) = 1 - 0.01

= P(Z < z ) = 0.99

= P(Z < 2.326 ) = 0. 99

z = 2.326

Using z-score formula,

x = z * +

x = 2.326 * 0.062 + 0.812

x = 0.96

x = 96%

b) P(Z < z ) = 0. 01

= P(Z < - 2.326 ) = 0. 01

z = - 2.326

Using z-score formula,

x = z * +

x = -2.326 * 0.062 +0.812

x = 0.67

x = 67%