Question

find the indicated probabilities using the geometric distribution, the poisson distribution, or the binomial distribution. then determine if the events are unusual. if convenient, use the appropriate probability table or technology to find the probabilities.
a newspaper finds that the mean number of typographical errors per page is eight. find the probability that (a) exactly five typographical errors are found on a page, (b) at most five typographical errors are found on a page, and (c) more than five typographical errors are found on a page.
(a) p(exactly five typographical errors are found on a page) = 0.0916
(round to four decimal places as needed.)
(b) p(at most five typographical errors are found on a page) = 0.1912
(round to four decimal places as needed.)
(c) p(more than five typographical errors are found on a page) =
(round to four decimal places as needed.)

Explanation:

Step1: Recall Poisson Probability Formula

The Poisson probability formula is \( P(X = k)=\frac{\lambda^{k}e^{-\lambda}}{k!} \), where \( \lambda \) is the mean, \( k \) is the number of events. Here, \( \lambda = 8 \), and we need \( P(X>5)=1 - P(X\leq5) \).

Step2: Calculate \( P(X\leq5) \)

We know from parts (a) and (b) that \( P(X = 5)=0.0916 \), and we can calculate or recall \( P(X = 0)+\cdots+P(X = 4) \), but since \( P(X\leq5)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)+P(X = 5) \). Alternatively, using the fact that \( P(X>5)=1 - P(X\leq5) \), and from part (b) \( P(X\leq5)=0.1912 \)? Wait, no, wait: Wait, part (b) is \( P(\text{at most five}) = 0.1912 \)? Wait, no, actually, let's recalculate. Wait, the mean \( \lambda = 8 \). Let's compute \( P(X\leq5) \) as the sum from \( k = 0 \) to \( k = 5 \) of \( \frac{8^{k}e^{-8}}{k!} \). But maybe from the problem, part (b) says \( P(\text{at most five})=0.1912 \)? Wait, no, the user's part (b) has \( 0.1912 \) as \( P(\text{at most five}) \). Then \( P(X>5)=1 - P(X\leq5)=1 - 0.1912 = 0.8088 \)? Wait, no, that can't be. Wait, maybe I misread. Wait, the mean is 8, so \( P(X\leq5) \) should be calculated correctly. Wait, let's use the Poisson formula. Let's compute \( P(X = 0)=\frac{8^{0}e^{-8}}{0!}=e^{-8}\approx0.000335 \)
\( P(X = 1)=\frac{8^{1}e^{-8}}{1!}=8e^{-8}\approx0.00268 \)
\( P(X = 2)=\frac{8^{2}e^{-8}}{2!}=\frac{64e^{-8}}{2}\approx0.01072 \)
\( P(X = 3)=\frac{8^{3}e^{-8}}{3!}=\frac{512e^{-8}}{6}\approx0.02869 \)
\( P(X = 4)=\frac{8^{4}e^{-8}}{4!}=\frac{4096e^{-8}}{24}\approx0.05739 \)
\( P(X = 5)=\frac{8^{5}e^{-8}}{5!}=\frac{32768e^{-8}}{120}\approx0.0918 \) (close to part (a)'s 0.0916, probably rounding differences). Now sum these: \( 0.000335+0.00268+0.01072+0.02869+0.05739+0.0918\approx0.1916 \), which is close to part (b)'s 0.1912 (rounding). Then \( P(X>5)=1 - 0.1912 = 0.8088 \)? Wait, but let's check with the formula. Alternatively, using the complement: \( P(X>5)=1 - P(X\leq5) \). If \( P(X\leq5)=0.1912 \), then \( P(X>5)=1 - 0.1912 = 0.8088 \). But let's verify with the Poisson distribution. The mean is 8, so the probabilities for \( k = 0 \) to \( k = 5 \) sum to approximately 0.1916, so subtracting from 1 gives approximately 0.8084, which rounds to 0.8088 (if part (b) is 0.1912). So we use \( P(X>5)=1 - P(X\leq5) \), where \( P(X\leq5)=0.1912 \) (from part (b)), so \( 1 - 0.1912 = 0.8088 \).

Answer:

\( 0.8088 \)