The manager of a computer help desk operation has collected enough data to conclude that the distribution of time per call is normally distributed with a mean equal to 8.21 minutes and a standard deviation of 2.14 minutes. what is the probability that three randomly monitored calls will each be completed in 4 minutes or less?

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.

only 10/60 isn't equivalent

hope this : )