Question

The number of M&M's in a package normally distributed with a mean of 48 candies and a standard deviation of 3. What is the number of candies that cuts off the top 5% of M&M's packages?

51.109

52.935

43.065

53.8799

Answer 1

Given that,

mean = = 48

standard deviation = =3

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z ) = 1 - 0.05

= P(Z < z ) = 0.95

z = 1.65 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.65 *3+48

x= 52.935