Question

AP-Stats-2005-Q2 2. Let the random variable X represent the number of telephone lines in use by the technical support center of a manufacturer at noon each day. The probability distribution of X is shown in the table below P(x) 0.35 0.20 0.15 0.15 0.10 0.05 ) Suppose you come by every day at noon to see how many lines are in use. What are the chances that you don't find all 5 in use until your 7" visit? ) Find the standard deviation of the number of lines in use this support center expects to have at noon. Each call lasts an average of 3.75 minutes. What is the mean and standard deviation of the number of minutes at noon? k) What are the mean and the standard deviation of the total number of lines in use this support center expects to have at noon over a 7-day week? l) m) Ifanother support center has a mean of 2.1 calls and a standard deviation of 1.8 call, what is the mean and standard deviation of the total of number of calls of both centers at noon?

Answer 1

Let Y no. of time that you find all 5 in use upto your 7th visit Then Y~ Binomial(7, 0.05) Required probability P(Y 0)1-0.05)0.6983 (j) E(X)0 0.35 10.202 (0.15) 3 (0.15) +4 * (0.10)5(0.05) E(X)-02 * 0.35 +12+0.20-22 *(0.15)+32 *(0.15) +42*(0.10)+52 *(0.05) Var(X) - E(X2) - E(x) 5 (1.6)2 2.44 SD(X)-V/Var(X) 1.5621 k. Z|X-no. of minutes the call lastsExponential distribution with mean = 3.751 E(Z|X) = 3.75X: E(Z) = 3.75E(X) = 3.75 * 1.6 6 Var(Z|X) = 3.752X: E(Var(Z|X)) = 3.752E(X) 3.752(Var(X) + E(x)) 3.752 * (2.44 +1.62) 70.3125 Var(E(Z|X)) 3.752Var(X) 3.752 * 2.44 = 34.3125 Var(Z)-E(Var(ZLX))+ Var(E(ZIX 70.3125 34.3125-104.625 SD(Z)-Var(Z) -10.2286 I. Required mean = 7E(X) = 7 * 1.6 = 11.2 Required SD- V/7Var(X)4.1328 m. Required mean 1.62.1- 3.7 Required sd-V2.441.82 2.3833