Question

the weights of boxes of a certain breakfast cereal are approximately normally distributed with a mean weight of 21 oz and a standard deviation of 0.06 oz. the lightest 5% of boxes do not meet minimum weight requirments and must be repackaged. to the nearest hundreth of an ounce, what is the minimum wegiht requirment for a cereal box?

Answer 1

solution:

mean weight of cereal box = oz

standard deviation = oz

probability of lightest boxes which require repackaging = 5% = 0.05

it is given that the weights are normally distributed

using normal table or z table, the value of z score with lightest 5% of weight box (to the left side of distribution ) is foundd to be -1.645

we know

so,

so the minimum weight required for a cereal box = 20.9 oz