Please could I have some help with the following question? My initial way of thinking was that Ui must be less than $5$ so that the measurement of the melting point is within $5 $ degrees of $c$, so I set up the inequality

$P(|U - 0n| < 5) ≥ \frac{1}{5^2 391n}$ , where 1/5^2 * 391n is equal to 90%.

However solving this to find n does not give me the integer answer the question asks for, so not sure what to do next.

"You want to determine the melting point c of a new material. You have n specimens on each of which you make a measurement of the melting point in degrees Kelvin, giving you a dataset m 1 ,…, m n . We model this with random variables M i =c+ U i , where U i is the random measurement error. It is known that E[ U i ]=0 and Var ( U i )=391 for each i , and that we may consider the random variables M 1 , M 2 ,… as independent. According to Chebyshev's inequality, how many measurements do you need to perform to be 90% sure that the average of the measurements is within 5 degrees of c ?"


$$Var\left(\frac{\sum_{i=1}^n M_i}{n} \right)=Var\left(\frac{\sum_{i=1}^n U_i}{n} \right)=\frac{Var(U_1)}{n}$$

$$P\left( \left|\frac{\sum_{i=1}^n M_i}{n}-c \right| \ge \frac{k\sqrt{Var(U_1)}}{n}\right) \le \frac1{k^2}$$

If $\epsilon = \frac{k\sqrt{Var(U_1)}}{n}$, then $k = \frac{\epsilon n}{\sqrt{Var(U_1)}}$. Hence ,

$$P\left( \left|\frac{\sum_{i=1}^n M_i}{n}-c \right| \ge \epsilon\right) \le \frac{Var(U_1)}{\epsilon^2n^2}$$

$$P\left( \left|\frac{\sum_{i=1}^n M_i}{n}-c \right| < \epsilon\right) \ge 1- \frac{Var(U_1)}{\epsilon^2n^2}$$

Hence, we want $$1- \frac{Var(U_1)}{\epsilon^2n^2} \ge 0.9$$

  • $\begingroup$ thank you! so plugging the numbers in would that be 1− 391 / 5 ^2 n ^2 ≥0.9 $\endgroup$ – Kiryne Dec 28 '18 at 13:15
  • $\begingroup$ Seems fine, just solve the inequality. $\endgroup$ – Siong Thye Goh Dec 28 '18 at 13:24
  • $\begingroup$ hmmm... what do you know about central limit theorem? $\endgroup$ – Siong Thye Goh Dec 28 '18 at 13:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.