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Question: The average number of misprints per page of a book is 1.5. Assuming the distribution of number of mi...

The average number of misprints per page of a book is 1.5. Assuming the distribution of number of misprints to be poisson, find the number of pages containing more than one misprint if the book contains 900 pages.

Answer

398

Explanation

Solution

Let XX be the number of misprints per page. The distribution of XX is given as Poisson with mean λ=1.5\lambda = 1.5. The probability mass function (PMF) of a Poisson distribution is P(X=k)=e−λλkk!P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} for k=0,1,2,…k = 0, 1, 2, \dots. Here, λ=1.5\lambda = 1.5, so P(X=k)=e−1.5(1.5)kk!P(X=k) = \frac{e^{-1.5} (1.5)^k}{k!}.

We need to find the number of pages containing more than one misprint. This corresponds to the event X>1X > 1. The probability of a page containing more than one misprint is P(X>1)P(X > 1). We can calculate this using the complement rule: P(X>1)=1−P(X≤1)P(X > 1) = 1 - P(X \le 1). P(X≤1)=P(X=0)+P(X=1)P(X \le 1) = P(X=0) + P(X=1).

Calculate P(X=0)P(X=0): P(X=0)=e−1.5(1.5)00!=e−1.5⋅11=e−1.5P(X=0) = \frac{e^{-1.5} (1.5)^0}{0!} = \frac{e^{-1.5} \cdot 1}{1} = e^{-1.5}.

Calculate P(X=1)P(X=1): P(X=1)=e−1.5(1.5)11!=e−1.5⋅1.51=1.5e−1.5P(X=1) = \frac{e^{-1.5} (1.5)^1}{1!} = \frac{e^{-1.5} \cdot 1.5}{1} = 1.5 e^{-1.5}.

Calculate P(X≤1)P(X \le 1): P(X≤1)=e−1.5+1.5e−1.5=(1+1.5)e−1.5=2.5e−1.5P(X \le 1) = e^{-1.5} + 1.5 e^{-1.5} = (1 + 1.5) e^{-1.5} = 2.5 e^{-1.5}.

Calculate P(X>1)P(X > 1): P(X>1)=1−P(X≤1)=1−2.5e−1.5P(X > 1) = 1 - P(X \le 1) = 1 - 2.5 e^{-1.5}.

The book contains 900 pages. The expected number of pages containing more than one misprint is the total number of pages multiplied by the probability that a single page contains more than one misprint. Number of pages with more than one misprint = Total pages ×P(X>1)\times P(X > 1) Number of pages with more than one misprint = 900×(1−2.5e−1.5)900 \times (1 - 2.5 e^{-1.5}).

To get a numerical answer, we need the value of e−1.5e^{-1.5}. Using e−1.5≈0.22313e^{-1.5} \approx 0.22313: Number of pages ≈900×(1−2.5×0.22313)\approx 900 \times (1 - 2.5 \times 0.22313) Number of pages ≈900×(1−0.557825)\approx 900 \times (1 - 0.557825) Number of pages ≈900×0.442175\approx 900 \times 0.442175 Number of pages ≈397.9575\approx 397.9575.

Since the number of pages must be an integer, we round the result to the nearest integer. Rounding 397.9575 gives 398.

The number of pages containing more than one misprint is approximately 398.