Answer to Question #300479 in Statistics and Probability for Kristel Tabil

Question

Solve the following problem:

A government office has six telephone lines. For the past months, the probability distribution of the random variable Y which represents the number busy line per day in shown in the records below

Y P(Y)

0 0.052

1 0.054

2 0.232

3 0.240

4 0.174

5 0.105

6 0.043

1. What is the probability that exactly four telephone

lines are busy in a day?

2.What is the probability that, at least ,four telephone lines busy in a day?

3.What is the probability that at least two but at most

four telephone lines busy in a day?

4.What is the probability that at least one telephone

lines are busy in a day?

5. What is the expected number of busy telephone lines in a day? Explain the results

6. What is the standard deviation of the number of busy telephone lines in a day? Explain the results.

Answer 1

1) it is given in the condition. $P(Y=4)=0.174$

2) $P(Y≥4)=P(Y=4)+P(Y=5)+P(Y=6)=0.174+0.105+0.043=0.322$

3)$P(2≤Y≤4)=P(Y=2)+P(Y=3)+P(Y=4)=0.232+0.240+0.174=0.646$

4)$P(Y≥ 1)=1-P(Y<1)=1-P(Y=0)=1-0.052=0.948$

5) $E(Y)=0*0.052+1*0.054+2*0.232+3*0.240+4*0.174+5*0.105+6*0.043=2.717$

This result means that in significant amount of time(month, for example), the mean number of busy lines per day will be around 2.717

6) $\sigma(Y)=\sqrt{E(Y^2)-E^2(Y)}=\sqrt{0^2*0.052+1^2*0.054+2^2*0.232+3^2*0.240+4^2*0.174+5^2*0.105+6^2*0.043-2.717^2}=1.65$

This result means that in significant amount of time(month, for example),we can expect that on around 68% of days the number of busy lines will deviate from the mean for no more than 1.65