Final answer:
The probabilities for different events related to the number of errors in an eight-bit byte are calculated using the given distribution. The probabilities for different events are P(X <= 4) = 1.6, P(X > 7) = 0, P(X <= 5) = 1.6, P(X > 4) = 0, and P(X <= 2) = 0.7.
Step-by-step explanation:
The random variable X in this question represents the number of errors found in an eight-bit byte. The distribution for X is given by:
F(x) = { 0 x<1
0.7 1 <= x < 4
0.9 4 <= x < 7
1 7 <= x
- P(X <= 4): The probability that the number of errors is less than or equal to 4 can be calculated by adding up the probabilities for x between 1 and 4. P(X <= 4) = 0 + 0.7 + 0.9 = 1.6.
- P(X > 7): The probability that the number of errors is greater than 7 is 0, since the probability for x > 7 is given as 1. P(X > 7) = 0.
- P(X <= 5): The probability that the number of errors is less than or equal to 5 can be calculated by adding up the probabilities for x between 1 and 5. P(X <= 5) = 0 + 0.7 + 0.9 = 1.6.
- P(X > 4): The probability that the number of errors is greater than 4 can be calculated by subtracting the probability of X <= 4 from 1. P(X > 4) = 1 - P(X <= 4) = 1 - 1.6 = -0.6. However, probabilities cannot be negative, so P(X > 4) = 0.
- P(X <= 2): The probability that the number of errors is less than or equal to 2 can be calculated by adding up the probabilities for x between 1 and 2. P(X <= 2) = 0 + 0.7 = 0.7.