Telephone call can be classified as voice (V) if someone is speaking, or data (D

Question

Telephone call can be classified as voice (V) if someone is speaking, or data (D) if there is a modem or fax transmission. Based on extension observation by the telephone company, we have the following probability model: P[V] = 0.65 and P[D] = 0.35. Assume that data calls and voice calls occur independently of one another, and define the random variable K_n to be the number of voice calls in a collection of n phone calls Compute the following. (a) E[K_100] = (b) sigma_K100 = Now use the central limit theorem to estimate the following probabilities. Since this is a discrete random variable, don't forget to use "continuity correction". (c) P[K_100 greaterthanorequalto 71] almostequalto (d) P[59 lessthanorequalto K_100 lessthanorequalto 78] almostequalto

Explanation / Answer

(a) E(K100) = np = 0.65 * 100 = 65

(b)  k100 = SQRT(NP(1-P) ) = sqrt( 100 * 0.65 * 0.35) = 4.77

(c) P (K100 >= 71) = ?

by Continuity correction factor table

P (K100 >= 71) = P ( K100 >= 71 -0.5) = P ( K100 >= 70.5)

Z - value = (70.5 - 65)/ 4.77 = 1.153

related p - value from z- table = 0.8749

so P (K100 >= 71) = 1 - 0.8749 = 0.1251

(d) P ( 59<= K 100 <= 78) =  P ( K100 =< 78) - P( K100 <= 59)

by Continuity correction factor table

P (K100 <= 59) = P ( K100 <= 59 - 0.5) = P ( K100 <= 58.5)

P (K100 =< 78) = P ( K100 =< 78 + 0.5) = P ( K100 >= 78.5)

so P ( 59<= K 100 <= 78) = P ( K100 =< 78.5) - P( K100 <= 58.5)

Respective z - values

for K100 = 78.5 => z = ( 78.5 - 65)/ 4.77 = 2.83 ; respective p- value= 0.9977

for K100 = 59.5 => z = ( 58.5 - 65)/ 4.77 = -1.36; respective p - value = 0.0869

so P ( 59<= K 100 <= 78) = 0.9977 - 0.0869 = 0.9108

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