Answer:
2000
Step-by-step explanation:
100 x 120 = 12000
12000 divided by 6 = 2000
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%20np%20%3D%202400%280.35%29%20%3D%20840%5C%5C%5Csigma%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B2400%280.35%29%281-0.35%29%7D%20%3D%20%24%2423.36)
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\%](https://tex.z-dn.net/?f=P%28800%20%5Cleq%20x%20%5Cleq%20850%29%20%3D%20P%28%5Cdisplaystyle%5Cfrac%7B800%20-%20840%7D%7B23.36%7D%20%5Cleq%20z%20%5Cleq%20%5Cdisplaystyle%5Cfrac%7B850-840%7D%7B23.36%7D%29%20%3D%20P%28-1.712%20%5Cleq%20z%20%5Cleq%200%29%5C%5C%5C%5C%3D%20P%28z%20%5Cleq%200.428%29%20-%20P%28z%20%3C%20-1.712%29%5C%5C%3D%200.666%20-%200.043%20%3D%200.623%20%3D%2062.3%5C%25)
![P(800 \leq x \leq 850) = 62.3\%](https://tex.z-dn.net/?f=P%28800%20%5Cleq%20x%20%5Cleq%20850%29%20%3D%2062.3%5C%25)
0.623 is the probability that between 800 and 850 files get damaged.