Answer:
Explanation:
a. To find the probability that two or fewer bits are transmitted in error, we need to add the probabilities of the events where X = 0, 1, or 2:
Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 0.55 + 0.22 + 0.13 = 0.90
Therefore, the probability that two or fewer bits are transmitted in error is 0.90.
b. To find the probability that at least one bit is transmitted in error, we need to subtract the probability that no bits are transmitted in error (i.e., X = 0) from 1:
Pr(X ≥ 1) = 1 - Pr(X = 0) = 1 - 0.55 = 0.45
Therefore, the probability that at least one bit is transmitted in error is 0.45.
c. To find the expected value of the number of bits transmitted in error, we need to multiply each possible value of X by its corresponding probability, and then sum the results:
E(X) = (0 × 0.55) + (1 × 0.22) + (2 × 0.13) + (3 × 0.08) + (4 × 0.02) = 0.79
Therefore, the expected value of the number of bits transmitted in error is 0.79.
d. To find the standard deviation of the number of bits transmitted in error, we first need to find the variance. We can do this using the formula:
Var(X) = E(X^2) - [E(X)]^2
To find E(X^2), we need to square each possible value of X, multiply it by its corresponding probability, and then sum the results:
E(X^2) = (0^2 × 0.55) + (1^2 × 0.22) + (2^2 × 0.13) + (3^2 × 0.08) + (4^2 × 0.02) = 0.99
Then, we can substitute the values we found into the formula for variance:
Var(X) = E(X^2) - [E(X)]^2 = 0.99 - 0.79^2 = 0.31
Finally, we can find the standard deviation by taking the square root of the variance:
SD(X) = √Var(X) = √0.31 ≈ 0.56
Therefore, the standard deviation of the number of bits transmitted in error is approximately 0.56.