Question

A typist claims that he prepares a typed page with typo errors of 1 per 10 pages. In a typing assignment of 40 pages, if the probability that the typo errors are at most 2 is p, then \(e^2 p =\)

• A. 5
• B. 13
• C. \(13e^{-2}\)
• D. \(5e^{-2}\)

Hint:

For Poisson distribution problems, always scale the mean \(\lambda\) linearly according to the size of the interval (here, the number of pages). If rate is \(r\) per unit, for \(n\) units, \(\lambda = n \times r\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The occurrence of typo errors on pages can be modeled using the Poisson Distribution. The probability mass function is given by \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \), where \(\lambda\) is the average rate of errors for the given interval (number of pages).
Step 2: Calculate the mean parameter \(\lambda\):
The rate of errors is given as "1 per 10 pages". For an assignment of 40 pages, the expected number of errors (\(\lambda\)) is: \[ \lambda = \frac{1}{10} \times 40 = 4 \]
Step 3: Calculate the probability \(p\):
We need the probability that the typo errors are at most 2, i.e., \( P(X \le 2) \). \[ p = P(X=0) + P(X=1) + P(X=2) \] Using the formula with \(\lambda = 4\): \[ p = \frac{e^{-4} \cdot 4^0}{0!} + \frac{e^{-4} \cdot 4^1}{1!} + \frac{e^{-4} \cdot 4^2}{2!} \] \[ p = e^{-4} \left( 1 + 4 + \frac{16}{2} \right) = e^{-4} (1 + 4 + 8) = 13e^{-4} \]
Step 4: Find the value of \(e^2 p\):
\[ e^2 p = e^2 \cdot (13e^{-4}) = 13e^{2-4} = 13e^{-2} \] Final Answer:
The value of \(e^2 p\) is \(13e^{-2}\).