Question

Suppose bit errors in a digital data file occur independently with probability p = 0.25 · 10^−6 per bit.

X = number of bit errors in a 1 Mbyte data file (= 10^6 bytes) Calculate exactly or with suitable approximation

(a) X's standard and standard deviation;

b) the probability that at least three bit errors occur in the data file.

c) Suppose that instead of an unknown parameter. A test file of size 3.85 Mbytes is checked where 10 bit errors are found. Enter a numerical point estimate of p and enter its Standard deviation d [ˆp].

Answer 1

Let, us Y be a Bernoulli random variable such that:

Here, we can denote them as follows:

P [Yi = 1]: The probability of occurrence of bit errors in i-th bit of a digital data file.

P [Yi = 0]: The probability of Non-occurrence of bit errors in i-th bit of a digital data file.

Where, P [Yi = 1] = p = 0.25 × 10−6

And P [Yi = 0] = 1- p = 1 - 0.25 × 10−6

Now, we know that 1 Mbyte data file = 106 bytes

We want to find the parameters of the Random variable X. which is nothing but summation of Yi’s. It can be written as:

So, X becomes a Binomial Random variable as X is the sum of 106 independent Bernoulli Random Variables. Where,

X ~ Bin (106, p)

Where, X can be described as the total number of bit errors in 1 Mbyte data file.

The P.m.f of X is given by:

  

[This transformation is obtained from the following Result:

          If Yi ~ Ber (p) independent and identically distributed where I =1(1)n

          And if

Then X ~ Bin (n, p)         ]

Now, we proceed with our problem:

(a) We have showed earlier that X ~ Bin (106, p)

So, Var(X) = 106 × p × (1-p)

                   = 106 ×   0.25 × 10−6 × (1 - 0.25 × 10−6)

                   ≈ 0.25

(b) Now we need to find the probability that at least three bit errors occur in the data file.

Mathematically we need to calculate:

P [at least three bit errors occur in the data file]

=P [X ≥ 3]

=1 - P [X < 3]

=1- {P [X = 0] + P [X = 1] + P [X = 2]}

= 1 – (0.778800759 + 0.19470019 + 0.024337512)

=0.00216154

(c) We know Point estimate for p is:

Where x is the number of times of the occurrences and n is the total number of trials.

Here x is given as the number of bit errors. (given as 10 in this case)

And n is the number of bits i.e. here the test file of size is 3.85 Mb = 3.85×106

So,

And point estimate of standard deviation is:

          ≈ 9.99997