Question

The marks in a Business Statistics class are normally distributed with a mean of 65% and a standard deviation of 12%. What percentage of the class received a mark of 90% or higher?

Answer 1

Solution:-

Given that,

mean = = 65% = 0.65

standard deviation = = 12% = 0.12

Using standard normal table,

P(Z > z) = 90%

= 1 - P(Z < z) = 0.90  

= P(Z < z) = 1 - 0.90

= P(Z < z ) = 0.10

= P(Z < 1.28 ) = 0.10  

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 0.12 + 0.65

x = 0.80