A new computer virus attacks a folder consisting of 200 files

[TomJerry]

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 - \mu/\sigma

For binomial distribution \mu = np and \sigma² = npq

Therefore
\mu = 40

\sigma = 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 ?

 

[CompuChip]

I think that you have made some minor mistakes.
First of all, npq is not the standard deviation, it is the variance (which is related to the st.dev. how?)

Moreover, this distribution is actually discrete and you are only using the normal distribution as an approximation. This means that you should probably apply the so-called continuity correction, by changing your upper bound a bit (it should not be 50 but ...?)

 

[TomJerry]

I think that you have made some minor mistakes.
First of all, npq is not the standard deviation, it is the variance (which is related to the st.dev. how?)

Moreover, this distribution is actually discrete and you are only using the normal distribution as an approximation. This means that you should probably apply the so-called continuity correction, by changing your upper bound a bit (it should not be 50 but ...?)

Thanks X should be 49 and not 50 ...Isnt that correct

 

[CompuChip]

How about 49.5?