Answer:
The probability that the amount dispensed per box will have to be increased is 0.0062.
Step-by-step explanation:
Consider the provided information.
Sample of 16 boxes is selected at random.
If the mean for 1 hour is 1 pound and the standard deviation is 0.1
1 Pound = 16 ounces , then 0.1 Pound = 16/10 = 1.6 ounces
Thus: μ = 16 ounces and σ = 1.6 ounces.
Compute the test statistic ![z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Cbar%20x-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z=\frac{15-16}{\frac{1.6}{\sqrt{16}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B15-16%7D%7B%5Cfrac%7B1.6%7D%7B%5Csqrt%7B16%7D%7D%7D)
![z=\frac{-1}{\frac{1.6}{4}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B-1%7D%7B%5Cfrac%7B1.6%7D%7B4%7D%7D)
![z=\frac{-1}{0.4}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B-1%7D%7B0.4%7D)
![z=-2.5](https://tex.z-dn.net/?f=z%3D-2.5)
By using the table.
P value = P(Z<-250) = 0.0062
Thus, the probability that the amount dispensed per box will have to be increased is 0.0062.