Question #98596
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?
1
Expert's answer
2019-11-13T11:58:38-0500

Let X=X= the weight of potato chips in ounces: XN(μ,σ2)X\sim N(\mu, \sigma^2)

Then


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

Given that μ0=10.2 ounces,σ=0.14 ounces\mu_0=10.2 \ ounces, \sigma=0.14\ ounces

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


P(X<10)=P(Z<10μnew0.14)=0.015P(X<10)=P(Z<{10-\mu_{new}\over 0.14})=0.015

Then


10μnew0.142.170090{10-\mu_{new}\over 0.14}\approx-2.170090

μnew=10+0.14(2.170090)10.3\mu_{new}=10+0.14(2.170090)\approx10.3

μnewμ0=10.310.2=0.1\mu_{new}-\mu_0=10.3-10.2=0.1

The  engineer needs to increase the mean by 0.10.1 ounces. New mean will be 10.310.3 ounces.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS