PROBABILITY SUMS OF RANDOM VARIABLES CENTRAL LIMIT THEOREM EXPECTED VALUES OF SU

Question

PROBABILITY
SUMS OF RANDOM VARIABLES
CENTRAL LIMIT THEOREM
EXPECTED VALUES OF SUMS
STANDARD DEVIATION

6.6.2 Telephone calls can be classified as voice (V) if someone is speaking, or data (D) if there is a mo- dem or fax transmission. Based on a lot of obser- vations taken by the telephone company, we have the following probability model: PIV] 0.8, PID] 0.2. Data calls and voice calls occur in- dependently of one another. The random variable Kn is the number of data calls in a collection of n phone calls. (a) What is E[K100l, the expected number of voice (b) What is 00, the standard deviation of the (c) Use the central limit theorem to estimate calls in a set of 100 calls? number of voice calls in a set of 100 calls? P[K1002 18], the probability of at least 18 voice calls in a set of 100 calls. (d) Use the central limit theorem to estimate P[16 16 and 24 voice calls in a set of 100 calls. K100 3 24], the probability of between

Explanation / Answer

Probability of being a Data call for a phone call, P(D) = 0.2. Phonecalls are independent. Then a phonecall is a Data call or not, can be described by a Bernoulli random variable with success probability P(D) = 0.2.

Now Kn is a random variable denoting number of data calls in a collection of n phone calls. Therefore, Kn is a Binomial random variable with parameters n and p=0.2.

(a) Then, K100 follows a Binomial (100, 0.2) distribution. Therefore, E(K100 ) = 100 x 0.2 = 20. If the expected number of voice calls are needed, then that will be 100 x 0.8 = 80.

(b) Standard dviation of number of data calls = sqrt(npq) = sqrt(100*0.2*0.8) = 4. Standard number of voice calls = sqrt(100*0.8*0.2) = 4, which is same as standard dev. of data calls.

(c) By CLT, we have for any sequence of random variable, the distribution of (Xn - E(Xn))/sigman follows a standard normal distribution. Then,

P[ K100 >= 18 ] = P [ (K100 - 20)/4 >= (17.5 - 20)/4 ] [Taking continuity correction]

= P( Z >= -0.625) = 0.734.

If the continuity correction is not done, then above form reduces to, P( Z >= -0.5 ) = 0.6915.

(d) Ignoring continuity correction we get the probability,

P [ 16 <= K100 <= 24] = P[ (16-20)/2 <= (K100 - 20)/2 <= (24-20)/2 ] = Phi(2) - Phi(-2) = 0.9545.

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