Answer to Question #224164 in Statistics and Probability for Begum

Question

(a) On average, a textbook author makes three word-processing errors per page on the first draft of her textbook. What is the probability that on the next page she will make (a) 5 or more errors? (b) no errors?

(b) A research scientist reports that mice will live an average of 40 months when their diets are sharply restricted and then enriched with vitamins and proteins. Assuming that the lifetimes of such mice are normally distributed with a standard deviation of 6.3 months, find the probability that a given mouse will live

i) more than 30 months; 

ii) less than 26 months; 

iii) between 35 and 47 months.

Answer 1

Solution:

(a):

Let 𝑋 be the number of errors made in one page. Then 𝑋 has a Poisson distribution with 𝜆 = 3 per page

(i) :$P(X\ge5)=1-P(X<5)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)]$

$=1-(\dfrac{e^{-3}3^0}{0!}+\dfrac{e^{-3}3^1}{1}+\dfrac{e^{-3}3^2}{2!}+\dfrac{e^{-3}3^3}{3!}+\dfrac{e^{-3}3^4}{4!}) \\=0.18474$

(ii): $P(X=0)=\dfrac{e^{-3}3^0}{0!}=0.04979$

(b):

$\mu=40,\sigma=6.3$

$X\sim N(\mu,\sigma)$

(i): $P(X>30)=1-P(X\le 30)=1-P(z\le \dfrac{30-40}{6.3})=1-P(z\le -1.59)$

$=1-[P(z\ge 1.59)] = P(z<1.59)=0.94408$

(ii): $P(X<26)=P(z\le \dfrac{26-40}{6.3})=P(z\le -2.22)=1-P(z<2.22)=0.01321$

(iii): $P(35\le X \le 47)=P(X\le 47)-P(X \le 35)=P(z\le \dfrac{47-40}{6.3})-P(z\le \dfrac{35-40}{6.3})$

$=P(z \le1.11)-P(z \le-0.79)$

$=P(z\le 1.11)-1+P(z\le 0.79) \\=0.86650-1+0.78524 \\=0.65174$