- #1
TomJerry
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Question:
A new computer virus attacks a folder consisting of 200 files. Each file gets damaged with probability 0.2 independently of other files. Using Normal approximation of binomial distribution, find the probability that fewer than 50 files get damaged.
Solution:
Here n=200, p =0.2, q=0.8.
Formulae for normal approx is Z = X - [tex]\mu[/tex]/[tex]\sigma[/tex]
For binomial distribution [tex]\mu[/tex] = np and [tex]\sigma[/tex]2 = npq
Therefore
[tex]\mu[/tex] = 40
[tex]\sigma[/tex] = 5.7
when X=50
Z = 50 - 40 / 5.7 = 1.8
P(X<50) = P(Z<1.8) = 0.5 - P(0<Z<1.8) = 0.5 - 0.4641 = 0.0359
Is this correct ?
A new computer virus attacks a folder consisting of 200 files. Each file gets damaged with probability 0.2 independently of other files. Using Normal approximation of binomial distribution, find the probability that fewer than 50 files get damaged.
Solution:
Here n=200, p =0.2, q=0.8.
Formulae for normal approx is Z = X - [tex]\mu[/tex]/[tex]\sigma[/tex]
For binomial distribution [tex]\mu[/tex] = np and [tex]\sigma[/tex]2 = npq
Therefore
[tex]\mu[/tex] = 40
[tex]\sigma[/tex] = 5.7
when X=50
Z = 50 - 40 / 5.7 = 1.8
P(X<50) = P(Z<1.8) = 0.5 - P(0<Z<1.8) = 0.5 - 0.4641 = 0.0359
Is this correct ?