Question

A virus that has a transmission rate of 42% hits a company with 258 employees. Let X be a random variable that counts the number of employees in the company that become infected with the virus.

A) What is the probability that 120 of the employees become infected with the virus?

B) What is the probability that less than 150 employees become infected with the virus?

C) What is the expected number of employees that will become infected with the virus?

D) What is the probability that more than 25% of the employees will become infected with the virus?

E) What is the probability that half or fewer of the employees will become infected with the virus?

Answer 1

Answer:-

Given that:-

A virus that has a transmission rate of 42% hits a company with 258 employees. Let X be a random variable that counts the number of employees in the company that become infected with the virus.

X be the no.of employes in the company that become infected with the virsus.

$X\sim Bin(258,0.42),$ where, n=258, p=0.42

$p(X=x)=\binom{258}{x}(0.42)^{x}(1-0.42)^{258-x}, x=0,1,2,....$

A) What is the probability that 120 of the employees become infected with the virus?

$p(x=120)=\binom{258}{120}(0.42)^{120}(0.58)^{258-120}$

$p(x=120)=0.0171$

B) What is the probability that less than 150 employees become infected with the virus?

$p(x<150)=p(x=0)+p(x=1)+....+p(x=149)$

  
(258 - (0.42) (0.58 ) 2014 (2) 0.42 0.05) 20.……. 258 10.42) (0.58) 149 (0.58) 258-
$=0.9999$

$p(x<150)\simeq 0.9999$

C) What is the expected number of employees that will become infected with the virus?

Expected Number =n.p=$258\times 0.42=108.3$

Expected Number $\simeq 108$

D) What is the probability that more than 25% of the employees will become infected with the virus?

25% empolyee$=258\times 0.25=64.5\simeq 65.$

$p(x>65)=1-(x\leq 65)$

  

= 1-pc = 0) + ...... + pl.x = 65)

$=1-[\binom{258}{0}(0.42)^{0}(0.58)^{258}+......+\binom{258}{65}(0.42)^{65}(0.58)^{258-65}]$$=1-0.000001$

$=0.999999$

$p(x>65)\simeq 0.9999$

E) What is the probability that half or fewer of the employees will become infected with the virus?

50% employee  $=258\times 0.5=129$

$p(x\leq 129)=p(x=0)+.................+p(x=129)$

  $=\binom{258}{0}(0.42)^{0}(0.58)^{258}+......+\binom{258}{129}(0.42)^{0}(0.58)^{258-129}$

$=0.9959$

$p(x\leq 129)=0.9959$

for further query please comment below.thank you