Answer to Question #199245 in Statistics and Probability for Anaya

Question

13. During a laboratory experiment, the average number of radioactive particles passing

through a counter in 1 millisecond is 4. What is the probability that 6 particles enter the

counter in a given millisecond?

14. The probability that a student at a local high school fails the screening test for scoliosis

(curvature of the spine) is known to be 0.004. Of the next 1875 students at the school

who are screened for scoliosis, find the probability that

(a) fewer than 5 fail the test;

(b) 8, 9, or 10 fail the test.

15. 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

(a) more than 32 months;

(b) less than 28 months;

(c) between 37 and 49 months.

Answer 1

13.) Assume that the particle stream is the easiest.

Then the number $\xi$ of radioactive particles passing through a counter in 1 millisecond has the Poisson's distribution, i.e.

$P(\xi =k)=\dfrac{\lambda^k}{k!}e^{- \lambda}$ where k=0,1,2,.... and by hypothesis $\lambda = 4$. So,

$P(\xi=6)=\dfrac{4^6}{6!}e^{-4}\approx 0.104$

14.) Let $X$ be the random variable representing the number of students who fail out of the next 1875 students.

If a student fails the screening test, we consider that as a success.

Then $𝑝 = 0.004.$ Trials are independent. Hence, $𝑋$ has a binomial distribution with parameters $𝑛 = 1875 \ and\ \ 𝑝 = 0.004$

$X$ ~$𝐵𝑖𝑛(𝑛, 𝑝), \ where \ \ 𝑛 = 1875 \ and \ \ 𝑝 = 0.004$

We have $𝑛 = 1875$ is large and $𝑝 = 0.004$ is near 0, then the binomial distribution can be approximated by the Poisson distribution with parameter 

$\mu=np=1875\times 0.004=7.5<10$

The Poisson distribution is a limiting case of the binomial distribution which arises when the number of trials 𝑛 increases indefinitely whilst the product $\mu=np$ , which is the expected value of the number of successes from the trials, remains constant.

Use Poisson distribution with p.m.f. $p(x;\mu)=\dfrac{e^{-\,u}\mu ^x}{x!}\ \ \ \ x=0,1,2,...$

(a) Fewer than 5 tails

$P(X<5)=P(X\leq 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$

$P(X<5)=\dfrac{e^{-7.5}7.5^0}{0!}+\dfrac{e^{-7.5}7.5^1}{1!}+\dfrac{e^{-7.5}7.5^2}{2!}+\dfrac{e^{-7.5}7.5^2}{3!}+\dfrac{e^{-7.5}7.5^4}{4!}$

$= 0.000553 + 0.004148 + 0.015555 + 0.038889 + 0.072916 \\=0.132061$

(b) 8, 9, or 10 fail the test 

$P(X=8)+P(X=9)+P(X=10)=\dfrac{e^{-7.5}7.5^8}{8!}+\dfrac{e^{-7.5}7.5^9}{9!}+\dfrac{e^{-7.5}7.5^{10}}{10!}$

$= 0.13733 + 0.11444 + 0.08583 = 0.33760$

15)

Mean =30 months

Standard Deviation= 6.3 months

Let x be the random variable for number of months

a) $P(X>32)$

$Z=\dfrac {X-mean} {sd}$

$=\dfrac {32-40} {6.3} ​$

$=-1.27$

$P(z>-1.27)=P(z<1.27)$ from the z tables is 0.89796

b) $P(X <28)$

$Z=\dfrac {28-40} {6.3}$

$=-1.90$

$P(z<-1.90)=1-P(z<1.90)$

P(z<1.90) from the z-tables is 0.97128

$P(z<-1.90)=1-0.97128 =0.02872$

c) $P(37<X<49)$

$Z_1=\dfrac {37-40} {6.3}$

$Z_2=\dfrac {49-40} {6.3} ​$

$=1.43$

$P(-0.48<z<1.43)=P(Z1<0.48)+ P(Z2<1.43)-1$

The values are obtained from the z tables

=0.68439+0.92364-1

=0.60803