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Pretoto4o

Answered question

2021-11-26

Complete the given table below by writing directly the needed data. Show complete solution in solving the Mean, Variance and Standard Deviation.

\begin{array}{cc} x & P (x) & x \cdot P (x) & x - μ & (x - μ)^{2} & (x - μ)^{2} P (x) \\ 0 & 0.11 \\ 1 & 0.22 \\ 2 & 0.33 \\ 3 & 0.25 \\ 4 & 0.09 \end{array}

Answer & Explanation

James Kilian

Beginner2021-11-27Added 20 answers

Step 1
Computation:

\begin{array}{cc} x & P (x) & x \cdot P (x) & x - μ & (x - μ)^{2} & (x - μ)^{2} P (x) \\ 0 & 0.11 & 0 & - 1.99 & 3.9601 & 0.43561 \\ 1 & 0.22 & 0.22 & - 0.99 & 0.9801 & 0.21562 \\ 2 & 0.33 & 0.66 & 0.01 & 0.0001 & 3.3 E - 05 \\ 3 & 0.25 & 0.75 & 1.01 & 1.0201 & 0.25503 \\ 4 & 0.09 & 0.36 & 2.01 & 4.0401 & 0.36361 \\ T o t a l & μ = \sum x P (x) = 1.99 & 0.05 & 10.0005 & 1.2699 \end{array}

From the table,
The value of mean is,

μ = \sum x P (x)

= 1.99

Step 2
The value of variance is,

σ^{2} = \sum {(x - μ)}^{2} P (x)

= 1.2699

The value of standard deviation is,

σ = \sqrt{σ^{2}}

= \sqrt{\sum {(x - μ)}^{2} P (x)}

= \sqrt{1.2699}

= 1.126898

\approx 1.1269

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