Question

The lifetime of a certain transistor in a certain application has mean 900 hours and standard deviation 30 hours. Find the mean and standard deviation of the length of time that four bulbs will last

Answer 1

Answer:

$E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4)$

$E(T) =900+900+900+900 =3600$

And the mean would be:

$\bar X = \frac{T}{4} = \frac{3600}{4}= 900$

And the standard deviation of total time would be:

$SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))$

$\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let total life time of t four transistors T = X1 + X2 + X3 + X4 (where X1,X2,X3 and X4 are life time of individual transistors

For this case the mean length of time that four transistors will last

$E(T)=E(X1 + X2 + X3 + X4 =E(x1) + E(x2) + E(x3) + E(x4)$

$E(T) =900+900+900+900 =3600$

And the mean would be:

$\bar X = \frac{T}{4} = \frac{3600}{4}= 900$

And the standard deviation of total time would be:

$SD(T)=\sqrt(Var(T)) = sqrt(Var(X1) + Var(x2) + Var(x3) + Var(x4))$

$\sigma=sqrt(30^2+30^2+30^2+30^2) =\sqrt(3600) =60$