Statistics Problems and Solutions Second Edition 2A.7 part (iii)

Question

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So I was looking back through Statistics Problems and Solutions (Second Edition) by Bassett et al., 1986 to see if there were any questions that I would consider revisiting to solve it again, and I found this question:

$2$A.$7\quad$The telephone exchange

A telephone receives, on average, $5$ calls per minute. Find the probability

(i) that in a $1$-minute period no calls are received;

(ii) that in a $2$-minute period fewer than $4$ calls are received;

(iii) that in a $20$-minute period no more than $102$ calls are received;

(iv) that out of five separate $1$-minute periods there are exactly four in which $2$ or more calls are received.

For this question, however, we will only be focusing on (iii) as it is the one on which I had the most trouble solving.

My attempt at solving

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What I know

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1. The number of calls $\mathit{X}$, will have a mean of $100$.
2. The solution requires that $\operatorname{Pr}(\mathit{X}\leq102)$, and since we know that the mean of the distribution is large, we can use normal approximation.
3. The appropriate normal distribution would be written as $\operatorname{\mathit{N}}(100, 10^2)$.

Now, since this is what we know, here is how I got my solution: $$\operatorname{Pr}(\mathit{X}\leq102)\approx\Phi\left(\frac{102+\frac{1}{2}-100}{10}\right)$$ $$ \begin{align} & =\Phi(0.25) \\ & \approx59.71\text{%} \end{align} $$ My question

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Is my solution correct, or how can I get the correct answer?

To clarify

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1. This is different from my last question here.
2. Sorry if my question seems short/trivial, or if I skipped any steps that would be necessary to solve the equation correctly.
3. I am only asking for a verification of my solution, and if I got it wrong, what could I do to attain the correct solution.

Answer 1

The arrival of calls can be modeled as a homogeneous Poisson process with intensity $\lambda = 5$ calls per minute. So in a span of $t$ minutes, $$X(t) \sim \operatorname{Poisson}(\lambda t)$$ counts the random number of calls observed. So in a span of $20$ minutes, the number of calls $X(20)$ is Poisson distributed with mean $20 \lambda = 100$ calls, and the exact desired probability is $$\Pr[X(20) \le 102] = \sum_{x=0}^{102} e^{-100} \frac{100^x}{x!}. \tag{1}$$ Since this sum has a large number of very small terms, it is not straightforward to calculate; we resort to a normal approximation. To do this correctly, we let $$Y \sim \operatorname{Normal}(\mu, \sigma^2)$$ such that our choice of mean $\mu$ and variance $\sigma^2$ matches the mean and variance of $X(20)$. This implies we must have $\mu = \lambda t = 100$, but since the variance of a Poisson random variable also equals its mean, $$\sigma^2 = \lambda t = 100.$$ So we have $$\Pr[X(20) \le 102] \approx \Pr[Y \le 102.5] = \Pr\left[\frac{Y - \mu}{\sigma} \le \frac{102.5 - 100}{10}\right] = \Pr[Z \le 0.25], \tag{2}$$ where $Z \sim \operatorname{Normal}(0,1)$ is standard normal and we have employed continuity correction. Hence the desired probability is approximately $\Phi(0.25) \approx 0.598706$.

Using a computer, the exact probability $(1)$ is $$0.60472101580283599044\ldots$$ so our approximation has an absolute error of about $0.006$, or a relative error of about $1\%$.

I'm not sure how you got your numerical answer $59.71\%$ but everything else was correct.