Question

certain computer virus can damage any file with probability 35%, independently of other files. Suppose this virus enters a folder containing 2400 files. Compute the probability that between 800 and 850 files get damaged (including 800 and 850).

Answer 1

Answer:

0.623 is the probability that between 800 and 850 files get damaged.

Step-by-step explanation:

We are given the following information:

We treat virus can damage computer as a success.

P( virus can damage computer) = 35% = 0.35

The conditions for normal distribution are satisfied.

By normal approximation:

$\mu = np = 2400(0.35) = 840\\\sigma = \sqrt{np(1-p)} = \sqrt{2400(0.35)(1-0.35)} = 23.36$

We have to evaluate probability that between 800 and 850 files get damaged.

$P(800 \leq x \leq 850) = P(\displaystyle\frac{800 - 840}{23.36} \leq z \leq \displaystyle\frac{850-840}{23.36}) = P(-1.712 \leq z \leq 0)\\\\= P(z \leq 0.428) - P(z < -1.712)\\= 0.666 - 0.043 = 0.623 = 62.3\%$

$P(800 \leq x \leq 850) = 62.3\%$

0.623 is the probability that between 800 and 850 files get damaged.