Answer:
1.41
Step-by-step explanation:
Given the table :
X : _____0 ___ 1 ____ 2 ____ 3 ____ 4 ____5 P(X= x)_ 11/48_7/24_ 3/16____1/6___ 1/12 __1/24
The standard deviation = √var(x)
The Variance : Var(X) = (Σx²*p(x)) - μ²
μ = E(X) = Σ(X * p(x)) :
Σ(X * p(x)) = (0*(11/48))+(1*(7/24))+(2*(3/16))+(3*(1/6))+(4*(1/12))+(5*(1/24))
μ = 1.70833333
Var(X): [(0^2*(11/48))+(1^2*(7/24))+(2^2*(3/16))+(3^2*(1/6))+(4^2*(1/12))+(5^2*(1/24))] - 1.70833333^2
= 4.9166667 - 2.9184027663889
Std (x) = √1.9982639336111
Std (x) = 1.4135996
Std (x) = 1.41
Answer:
The Standard deviation of the random variable is 1.414.
Step-by-step explanation:
We Know,
Standard deviation= [tex]\sqrt{variance}[/tex]
Variance is given by:
[tex]Var(X)=E(X^{2} )-E(X)^{2}\\[/tex]
Given,
[tex]x = 0\ 1\ 2\ 3\ 4\ 5\\P(X=x)= \frac{11}{48} \ \frac{7}{24} \ \frac{3}{16}\ \frac{1}{6}\ \frac{1}{12}\ \frac{1}{24} \\[/tex]
[tex]E(X)=[/tex]∑ [tex]x*P(X)[/tex]
[tex]E(X) = 0+ 1*\frac{7}{24} + 2*\frac{3}{16} + 3*\frac{1}{6} + 4*\frac{1}{12} + 5*\frac{1}{24}\\= 1.7083[/tex]
[tex]E(X^{2} ) =[/tex][tex]0+ 1^{2} *\frac{7}{24} + 2^{2} *\frac{3}{16} + 3^{2} *\frac{1}{6} + 4^{2} *\frac{1}{12} + 5^{2} *\frac{1}{24}[/tex]
[tex]= 4.9167[/tex]
So,
[tex]Var(X)= 4.9167- 1.7083^{2}[/tex]
[tex]= 1.99841\\\\Hence, Standard Deviation= \sqrt{1.99841} \\\\ Standard Deviation= 1.413( (upto 3 decimals)[/tex]
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