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Airida [17]
3 years ago
13

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
Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0

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

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3 years ago
An element with mass 590 grams decays by 19.5% per minute. How much of the
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Step-by-step explanation:

Step 1:

Given,

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Step 2 :

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4 years ago
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3 years ago
What is 1 1/2 subtracted from the sum of 4 2/3 and 5 2/5
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Hi there!

First, we need to add 4 2/3 and 5 2/5.

We should begin with converting the fractions into improper fractions.
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Hope this helps!
3 0
3 years ago
Read 2 more answers
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