Answer:
0.0003
Step-by-step explanation:
Mean=μ=8.21
Standard deviation=σ=2.14
We have to find P(3 randomly monitored call completed in 4 min or less).
P(Xbar≤4)=?
μxbar=μ=8.21
σxbar=σ/√n=2.14/√3=1.2355
Z-score associated with xbar=4
Z=[Xbar-μxbar]/σxbar
Z=[4-8.21]/1.2355
Z=-4.21/1.2355
Z=-3.4075
P(Xbar≤4)=P(Z≤-3.41)
P(Xbar≤4)=P(-∞<Z<0)-P(0<Z<-3.41)
P(Xbar≤4)=0.5-0.4997
P(Xbar≤4)=0.0003
Thus, the probability that three randomly monitored calls will each be completed in 4 minutes or less is 0.0003.