A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only 2% of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must m equal? Assume the filling weights have a normal distribution. Anderson, David R.. Essentials of Statistics for Business and Economics (p. 299). South-Western College Pub. Kindle Edition.

Answer:The mean of weight is 30.3Step-by-step explanation:We are given the following information in the question:Standard Deviation, σ = 6 ouncesWe are given that the weights is a bell shaped distribution that is a normal distribution.Formula:[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]P(container holds less than 18 ounces) = 2% = 0.02P(x < 18) = 0.02[tex]P( x < 18) = P( z < \displaystyle\frac{18 - \mu}{6}) = 0.02[/tex]Calculation the value from standard normal z table, we have,  P(Z < -2.05) = 0.02[tex]\displaystyle\frac{18 - \mu}{6} = -2.05\\\\\mu =18- (6\times -2.05) = 30.3[/tex]Hence, the mean of weight is 30.3