A machine fills tins with paint. The mass of a tin is normally distributed with a mean of 0.3 kg and a standard deviation of 0.02 kg. The mass of the paint that goes into the tin is normally distributed with a mean of 10 kg and a standard deviation of 0.05 kg.

The Quality Control department recommends that the average mass of 4 randomly selected filled tins must lie between 10.20 and 10.40 kg with 99.9% probability.

Question : What is the probability that the mass of a single filled tin lies between 10.20 and 10.40 kg?

Answer : This is what I have done,

X~N(0.3, 0.0004), where X represents the mass of tin

Y~N(10, 0.0025), where Y represents the mass of paint

Let U = X+Y. Since both are independent, U~(10.3, 0.0029)

By applying the Central Limit Theorem,

P(10.2 < U < 10.4) = P( $\left(\frac{10.2 - 4(10.3)}{\sqrt{4*0.0029}}\right)$<$\left(\frac{U - 4(10.3)}{\sqrt{4*0.0029}}\right)$<$\left(\frac{10.4 - 4(10.3)}{\sqrt{4*0.0029}}\right)$ )

However, this would result in large values of CDF where I wouldn't be able to use the table for my answer.

Can anyone tell me where I went wrong? Thanks~

  • 1
    $\begingroup$ The question is about a single tin. Why do you muliply the variance and the expected value by $4$? $\endgroup$ – callculus May 17 '18 at 13:01
  • 1
    $\begingroup$ @callculus ah thats true, thanks for the help! $\endgroup$ – Wei Xiong Yeo May 17 '18 at 13:02
  • $\begingroup$ You´re welcome. If you have additional questions about this exercise feel free to ask. $\endgroup$ – callculus May 17 '18 at 13:13
  • 1
    $\begingroup$ Another hint: $P(-a<Z<a)=2\Phi(a)-1$, where $Z \sim \mathcal N(0,1)$ $\endgroup$ – callculus May 17 '18 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.