Answer :
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.
The probability will be "0.0003".
Probability:
In some kind of a randomized experimentation, probability would be measure of the possibility that an outcome will happen. The greater the likelihood of such an occurrence, the further probable it will materialize.
According to the question,
Mean, μ or [tex]\mu \bar x[/tex] = 8.21
Standard deviation, σ = 2.14
[tex]\sigma \bar x[/tex] = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
= [tex]\frac{2.14}{\sqrt{3} }[/tex]
= 1.2355
Now,
Z-score will be:
= [tex]\frac{\bar X- \mu \bar x}{\sigma \bar x}[/tex]
By substituting the values,
= [tex]\frac{4-8.21}{1.2355}[/tex]
= [tex]-\frac{4.21}{1.2355}[/tex]
= -3.4075
hence,
The probability:
P ([tex]\bar X \leq 4[/tex]) = P (Z [tex]\leq[/tex] -3.41)
= P (-∞ < Z < 0) - P(0 < Z < -3.14)
= 0.5 - 0.4997
= 0.0003
Thus the above answer is correct.
Find out more information about probability here:
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