$\therefore \rm E\left[ {{Y^2}\left( t \right)} \right] = \frac{K}{{2RC}}$
12. Let X be a real-valued random variable with E[X] and E[X2] denoting the mean values of X and X2, respectively. The relation which always holds is
(E[X])2 > E(X2)
E(X2) ≥ (E[X])2
E[X2] = (E[X])2
E[X2] > (E[X])2
Answer.2. E(X2) ≥ (E[X])2
Explanation
E[X] denotes the mean value of ‘X’
E[X2] denotes the mean value of X2
Also, the variance of Random variable ‘X’ is given as:
Var(X) = E(X2) – [E(X)]2
Note: Variance is always non-negative, i.e.
E[X2] – [E(X)]2 ≥ 0
∴ E[X2] ≥ (E[X])2
13. A digital communication system transmits a block of N bits. The probability of error in decoding a bit is α. The error event of each bit is independent of the error events of the other bits. The received block is declared erroneous if at least one of its bits is decoded wrongly. The probability that the received block is erroneous is
N(1 – α)
αN
1 – αN
1 – (1 – α)N
Answer.4. 1 – (1 – α)N
Explanation
Since all the bits are independent of each other, the probability that all the bits are received
correctly (PC) will be:
PC = (1 − α)1 × (1 − α)2 × _ _ _ _ _(1 − α)N
PC = (1 − α)N
where α is the probability of error in detection.
Now, the probability that the received block is erroneous will be:
Pe = 1 – PC
Pe = 1 – (1 – α)N
14. Find median and mode of the messages received on 9 consecutive days 15, 11, 9, 5, 18, 4, 15, 13, 17.
13, 6
13, 18
18, 15
15, 16
Answer.2. 13, 18
Explanation
Arranging the terms in ascending order 4, 5, 9, 11, 13, 14, 15, 18, 18.
Median is (n + 1)/2 term as n = 9 (odd) = 13.
Mode = 18 which is repeated twice.
15. Bits 1 and 0 are transmitted with equal probability. At the receiver, the probability density function of the respective received signals for both bit are as shown below
1/2
1/4
1/16
1/32
Answer.3. 1/16
Explanation
The received signal is said to an error signal if we receive 1 but transmitted signal was 0, or if we receive 0 but transmitted signal was 1.
17. A coin is tossed up 4 times. The probability that tails turn up in 3 cases is ______
1/2
1/3
1/4
1/6
Answer.1. 1/2
Explanation
p = 0.5 (Probability of tail)
q = 1 − 0.5 = 0.5
n = 4 and x is binomial variate.
P (X = x) = nCx px qn-x.
P (X = 3) = 4C3 (0.5)3 = 1⁄2.
18. Which of the following statistical method can be used for a single sample data?
Frequency distribution
Uncertainty distribution
Standard deviation
None of the above
Answer.1. Frequency distribution
Explanation
In statistics, a frequency distribution is a list, table, or graph that displays the frequency of various outcomes in a sample.
The frequency of observation tells you the number of times the observation occurs in the data.
Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval.
Discrete data is generated by counting each and every observation.
When an observation is repeated, it is counted. The number for which the observation is repeated is called the frequency of that observation.
The class limits in discrete data are true class limits, there are no class boundaries in discrete data.
19. If E denotes the expectation the variance of a random variable X is denoted as?
(E(X))2
E(X2) − (E(X))2
E(X2)
2E(X)
Answer.2. E(X2) − (E(X))2
Explanation
This property of the mathematical expectation states that if there is an X and Y, then the product of those two random variables is equal to the product of the mathematical expectation of the individual random variables. In other words, E(X2) − (E(X))2, provided all the expectations exist.
20. Let X ∈ {0,1} and Y ∈ {0,1} be two independent binary random variables. If P(X = 0) = p and P(Y = 0) = q, then P(X + Y) ≥ 1 is equal to
pq + (1 – p)(1 – q)
pq
p(1 – q)
1 – pq
Answer.4. 1 – pq
Explanation
1) P(X = 0) + P(X = 1) = 1
2) If Z = X + Y, then pdf (Z) = convolution of pdf (x) and pdf (y)
21. The spectral density and autocorrelation function of white noise is, respectively;
Delta and Uniform
Uniform and Delta
Gaussian and Uniform
Gaussian and Delta
Answer.2. Uniform and Delta
Explanation
The spectral density of white noise is Uniform and the autocorrelation function of White noise is the Delta function. A white noise process is a random process of random variables that are uncorrelated, have mean zero, and a finite variance.
22. The random variables X and Y have variances 0.2 and 0.5 respectively. Let Z= 5X − 2Y. The variance of Z is?