Question

Suppose a professor for organic chemistry 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 73.5% and standard deviation 9.8%.

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

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

A) 99.8% B) 53.2%

A) 96.3% B) 50.7%

A) 89.1% B) 48.5%

Answer 1

Solution:-

Given that,

mean = = 73.5%

standard deviation = = 9.8%.

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 < 2.33) = 0.99

z = 2.33 ( using z table )

Using z-score formula,

x = z * +

x =2.33 * 9.8%.+73.5%

x = 96.3%

b.

Using standard normal table,

P(Z < z) = 0.01

= P(Z < z) =0.01

z = -2.33 ( using z table )

Using z-score formula,

x = z * +

x =-2.33 * 9.8%.+73.5%

x = 50.7%

A) 96.3% B) 50.7%