Answer to Question #98596 in Statistics and Probability for kiyan

Question

The weight of potato chips in a medium‐size bag is stated to be 10 ounces. The amount that the packaging machine puts in these bags is believed to have a Normal distribution with a mean of 10.2 ounces and a standard deviation of 0.14 ounces.
1.If the production engineer wants to change the value of the mean but maintain the standard deviation at 0.14 ounces so that at most 1.5% of all bags are underweight, how large does the engineer need to make that mean?

Answer 1

Let $X=$ the weight of potato chips in ounces: $X\sim N(\mu, \sigma^2)$

Then

$$Z={X-\mu\over \sigma}\sim N(0,1)$$

Given that $\mu_0=10.2 \ ounces, \sigma=0.14\ ounces$

If $P(X<10)=0.015,$ find $\mu_{new}$

$$P(X<10)=P(Z<{10-\mu_{new}\over 0.14})=0.015$$

Then

$${10-\mu_{new}\over 0.14}\approx-2.170090$$

$$\mu_{new}=10+0.14(2.170090)\approx10.3$$

$$\mu_{new}-\mu_0=10.3-10.2=0.1$$

The  engineer needs to increase the mean by $0.1$ ounces. New mean will be $10.3$ ounces.