Answer:
a)Probability( exactly 18 will have a useful life of at least 500 hours
)= 0.1368
b) Probability(at least 15 will have a useful life of at least 500 hours)=0.8042
Step-by-step explanation:
Let X be random variable.It represent the number of light having a usefull life of at least 500 hours 20 fluorescent lights
Lets consider it success if a light has a useful life of at least 500 hours . so
Probability of a success in each trial is p =0.8
Because of the trials are independent, X has Binomial distribution with parameters n= 20 and p=0.8
The probability mass function is,
P(X=x)=b(x;20,0.8)
where x=0,1,2,3,4....20
=[tex]\left(\begin{array}{c}{20} \\ {x}\end{array}\right)(0.08)^x(1-0.80)^(20-x)[/tex]
[tex]=\left(\begin{array}{c}{20} \\ {x}\end{array}\right)(0.08)^x(0.20)^{(20-x)}[/tex]................(1)
A)exactly 18 will have a useful life of at least 500 hours
We Have to find the probability that exactly 18 will have a useful life of at least 500 hours.
using equation (1)
P(X= 18) =b(18,20,0.08)
P(X= 18)=[tex]\left(\begin{array}{c}{20} \\ {x}\end{array}\right)(0.80)^18(0.20)^{(20-18)}[/tex]
P(X= 18)=[tex]\frac{20!}{18!2!}(0.80)^{18}(0.20)^2[/tex]
P(X= 18)=[tex]190\times 0.0180 \times 0.04[/tex]
P(X= 18)=0.1368
B) at least 15 will have a useful life of at least 500 hours
We Have to find the probability that at least 15 will have a useful life of at least 500 hours
P(X>= 15) = 1-P(X<=14)
P(X>= 15)=1-[tex]\sum_{x=0}^{4} b(x ; 20,0.80)[/tex]
P(X>= 15) = [tex]1-[\frac{20!}{14!6!}(0.80)^{14}(0.20)^6+\frac{20!}{13!7!}(0.80)^{13}(0.20)^7+............+\frac{20!}{1!19!}(0.80)^{1}(0.20)^{19} +\frac{20!}{0!20!}(0.80)^{0}(0.20)^{20}][/tex]
P(X>= 15)=[tex]1-[0.109+0.0545+0.0222+0.0074+0.0020+0.0005+0.0001+0+0+0+0+0+0+0+0][/tex]
P(X>= 15)1= -0.1958
P(X>= 15)=0.8042
