A new computer virus attacks a folder consisting of 200 files. Each file gets damaged with probability 0.2 independently of other files. What is the probability that fewer than 50 files will get damaged? g

0.95345

Explanation:

Given that :

Number of files (n) = 200

Probability of getting damaged (p) = 0.2

q = 1 - p = 1 - 0.2 = 0.8

using normal approximation:

Mean (m) = np = 200 * 0.2 = 40

Standard deviation (s) = √n*p*q= √(200 * 0.2 * 0.8) = 5.6568542

P(X < 50)

X = 50 - 0.5 = 49.5

Find the standardized score :

Z = (x - m) / s

Z = (49.5 - 40) / 5.6568542

Z = 9.5 / 5.6568542

Zscore = 1.6793786

P(Z < 1.6793) = 0.95345 ( using Z probability calculator)

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