The random variable X is the number of siblings of a student selected at random from a particular secondary school. Its probability distribution is given in the table. Find the standard deviation of the random variable. Round to three decimal places. x 0 1 2 3 4 5 P(X= x) 11 7 3 1 1 1 48 24 16 6 12 24

The Standard deviation of the random variable is 1.414.

Step-by-step explanation:

We Know,

Standard deviation=

Variance is given by:

Given,

∑

So,

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1.41

Step-by-step explanation:

Given the table :

X : 0 ___ 1 2 3 4 5 P(X= x)_ 11/48_7/24_ 3/161/6___ 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

Month                 No. of orders in         No. of orders out
January (1)             3                                     1
February (2)           4                                     3
March (3)                5                                      5
April (4)                   6                                      7

Number of orders in = number of month + 2

f(x) = x + 2.

Number of orders out = 2 number of month + 1

g(x) = 2x - 1

So, in words, the numbers of orders in is equal to the number of month plus 1. This represents a linear function, with slope 1 and y-intercept 2. And the numbers of orders out is also a linear function with slope 2 and y-intercpet - 1.