Question

Question 13: The average number of misprints per page of a book is 1.5.Assuming the distribution of number of misprints to be Poisson. The probability that a particular book is free from misprints,is B. 0.435 D. 0.2231 A. 0.329 C. 0.549​

Answer 1

Answer:

P(0≤X≤5)≈P(−0.5≤X≤5.5)

=P(−0.5−1010√≤Z≤5.5−1010√)

=P(−3.32≤Z≤−1.42)

=P(Z≤−1.42)−P(Z≤−3.32)

=0.0778−0.005

=0.0728.

So the approximate probability is 0.0728.

Getting the probability from the “exact” distribution which in this case is the Poisson, I get the following:

P(0≤X≤5)=∑5i=0(10)ie−10i!

≈0.067.

Answer 2

Answer:

The probability that a particular book is free from misprints is option D (0.2231)

Step-by-step explanation:

• Poisson Distribution is a discrete frequency distribution that gives the probability of a number of independent events occurring in a fixed time.
• Formulae for the Poisson Distribution is given by

        $P=\frac{e^{-m}*m^{r} }{r!}$

         Where m is the mean of the distribution and r is Poisson random

         variable

Step 1 of 1:

• Given m=1.5, r=0
• The probability of the distribution is

           $P=\frac{e^{-m}*m^{r} }{r!}$

           $P=\frac{e^{-1.5}*1.5^{0} }{0!}$

         $P=\frac{e^{-1.5}*1 }{1}\\P=0.2231$

Conclusion:

The probability that a particular book is free from misprints is 0.2231