A team of 3 QA testers and l QA team lead is working on a project. The team finishes

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The team-lead gets $2000 if the team of 3 QA testers and l QA team lead is working on a project. The team finishes the job and is paid.

<h3>What is average?</h3>

It is defined as the single number that represents the mean value for the given set of data or the closed value for each entry given in the set of data.

Let's suppose the team lead salary is x and the team-lead receives, $600 more than the average salary of the entire team then,

The team lead salary:

$\rm x = \dfrac{1200+1200+1200+x}{4} + 600$

4x = 3600+ x+ 2400

3x = 6000

x = $2000

Thus, the team lead gets $2000 if the team of 3 QA testers and l QA team lead is working on a project. The team finishes the job and is paid.

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Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$