Question

A machine dispenses potato chips into bags that are advertised as containing one pound of product. To be on the safe side, the machine is supposed to be calibrated to dispense 16.07 ounces per bag, and from long time observation, the distribution of the fill-weights is known to be approximately normal and the process is known to have a standard deviation of 0.15 ounces.

If the machine is properly calibrated, what proportions of bags will be underweight?

Answer 1

Here, μ = 16.07, σ = 0.15 and x = 16. We need to compute P(X <= 16). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (16 - 16.07)/0.15 = -0.47

Therefore,
P(X <= 16) = P(z <= (16 - 16.07)/0.15)
= P(z <= -0.47)
= 0.3192

proportion of underweight bags = 0.3192