Question

A company has a juice dispensing machine that dispenses orange juice into 10 oz bottles. The distribution for the amount of juice dispensed by the machine follows a normal distribution with a standard deviation of 0.12 ounce. The company can control the mean amount of juice dispensed by the machine.

What value of the mean should the company use if it wants to guarantee that 98.75% of the bottles contain at least 10 ounces (the amount on the label)?

Answer 1

We have to calculate such that

P(X > x) = 0.9875

We convert this to standard normal as

P(Z > ( x - ) / ) = 0.9875

P(Z < ( x - ) / ) = 1 - 0.9875

P(Z < ( x - ) / ) = 0.0125

From Z table, z-score for the probability of 0.0125 is -2.2414

( x - ) / = -2.2414

( 10 - ) / 0.12 = -2.2414

= 10.27