Answer and Step-by-step explanation:
From the question statement we get know that it is Binomial distribution because there are only two possible outcomes so we need to use Binomial Probability Distribution for this question.
Formula for the Binomial Probability Distribution:
P(X)= p^x q^(n-x)
Where,
- C_x^n=n!/(n-x)!x!
(i.e. combination)
- x= total number of successes
- p=probability of success (p=1-q)
- q=probability of failure (q=1-p)
- n=number of trials
- P(X)= probability of total number of successes
Answer and explanation for each part of the question are as follow:
a.What is the probability that among 10 randomly observed individuals exactly 4 do not cover their mouth when sneezing?
Solution:
Given that
n=10
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=4 (number of successes i.e. individuals not covering their mouths)
C_x^n=n!/(n-x)!x!=10!/(10-4)!4!=210
P(X)=C_x^n p^x q^(n-x)=210×〖(0.267)〗^4×〖0.733〗^(10-4)
P(X)=210×0.00508×0.155
P(X)=0.165465
b. What is the probability that among 10 randomly observed individuals fewer than 3 do not cover their mouth when sneezing?
Solution:
Given that
n=10
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=3 (number of successes i.e. individuals not covering their mouths)
C_x^n=n!/(n-x)!x!=10!/(10-3)!3!=120
P(X)=C_x^n p^x q^(n-x)=120×(0.267)^3×〖0.733〗^(10-3)
P(X)=120×0.01903×0.1136
P(X)=0.25962
c. Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? why?
Solution:
Given that
n=18
p=0.267 (because p is the probability of success which is “number of individuals not covering their mouths when sneezing” in the question)
q=1-0.267=0.733
x=9 (x is the number of successes “number of individuals not covering their mouths when sneezing”, if less than half cover their mouth then more than half will not cover), so let x=9
C_x^n=n!/(n-x)!x!=18!/(18-9)!9!=48620
P(X)=48620×(0.267)^9×〖0.733〗^(18-9)
P(X)=48620×0.00000689×0.0610
P(X)=0.020
Yes, I am surprised that probability of less than 9 individuals covering their mouth when sneezing is 0.020. Which is extremely is small.
, the contribution of the terms of degree less than 2 becomes negligible, which means we can write
![\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto1%7D%5Cfrac%7B%5Csqrt%5B3%5Dx-x%7D%7B%5Csqrt%20x-x%7D%3D%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bx%5E%7B1%2F3%7D-x%7D%7Bx%5E%7B1%2F2%7D-x%7D)
and 
, we can simplify the first term:
