In a hotel, large number of cups and saucers are washed everyday. The number of cups that are broken each day while washing averages $2.1$. The number of saucers broken each day averages $1.6$, independently of the number of cups broken. A random sample of $100$ weeks is chosen. Using a suitable approximation, find the probability that the average weekly total number of broken cups and saucers does not exceed $26$.

Define $T$ to be the total number of cups and saucers broken during a week of $7$ days. Then $T \sim Po(7 \times 2.1+7\times 1.6=25.9)$

There are two ways to find the required probability.

First Method:

$P(T_1+T_2+...+T_{100} \leq 2600)$. Since $\lambda>10$, by normal approximation, the distribution $$Y=\sum_{i=1}^{100}{T_i} \sim N(2590,2590)$$ Hence, the required probability is $P(T_1+T_2+...+T_{100} \leq 2600)=P(Y < 2600.5)=0.578$

Second Method:

Since $n$ is large, by Central Limit Theorem, we have $$\bar{T} \sim N(25.9, \frac{25.9}{100})$$ Hence, the required probability is $P(\bar{T}<26)=0.578$

Question: Why we don't need to use continuity correction in the second method? I thought as long as we use approximation from discrete r.v. to normal r.v., we need to perform continuity correction. But in this case, it is not.

Can anyone explain to me?

  • $\begingroup$ What do you mean by $\sim (2590,2590)$ and $\sim (25.9, \frac{25.9}{100})$ ? $\endgroup$ – callculus Sep 19 '14 at 7:58
  • $\begingroup$ @calculus: Thanks for pointing out. Edited. $\endgroup$ – Idonknow Sep 19 '14 at 9:27

The answer is yes and no: You only really need to use it if the area under the normal curve will change substantially with a 0.5 unit change in the location on the x axis. In the normal approximation, your mean and variance are large, so it is not necessary.

In the second case, the underlying distribution is not defined over the integers, but over a discrete, finite set of fractions much smaller than the integers, hence again the proper correction in that case will not affect the answer.


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