Respuesta :
The sum of the variance of the 4 dots is equal to the variance of the sum of
the masses of the 4 dots.
- The correct option for standard deviation of T is; [tex]\underline{\sigma_T = 0.5 \ grams}[/tex]
Reasons:
The given parameter are;
The mean mass of the dots of frosting, μ = 3 grams
The standard deviation of each dot of frosting, σ = 0.25 grams
The total mass of the 4 dots = T
Required:
To determine the standard deviation of T
Solution:
Standard deviation, σ = √(Variance)
Therefore;
Variance, Var = σ²
The sum of the variance of a given number of independent random variable is presented as follows;
- [tex]\displaystyle Var \left(\sum_{i=1}^{n} X_i \right) = \mathbf{ \sum_{i = 1}^{n}Var(X_i)}[/tex]
The variance of each sample, are; Var(X) = σ² = 0.25²
Therefore, for the four samples, we have;
Var(X₁) = Var(X₂) = Var(X₃) = Var(X₄) = σ² = 0.25²
Which gives;
[tex]\displaystyle \mathbf{Var \left(\sum_{i=1}^{4} X_i \right)} = \sum_{i = 1}^{4}Var(X_i) = Var(X_1) + Var(X_2) +Var(X_3) +Var(X_4)[/tex]
Var(X₁) + Var(X₂) + Var(X₃) + Var(X₄) = 0.25² + 0.25² + 0.25² + 0.25² = 0.25
Therefore;
[tex]\displaystyle Var \left(\sum_{i=1}^{4} X_i \right) =\sigma^2_{1 + 2 + 3 +4} = \sum_{i = 1}^{4}Var(X_i) = \mathbf{0.25}[/tex]
[tex]\displaystyle \sigma^2_T = \sigma^2_{1 + 2 + 3 +4}[/tex]
[tex]\displaystyle Var \left(\sum_{i=1}^{4} X_i \right) =\sigma^2_T = \sum_{i = 1}^{4}Var(X_i) = 0.25[/tex]
[tex]\displaystyle Standard \ deviation \ of \ the \ total \ mass = \sigma_T = \sqrt{ \sum_{i = 1}^{4}Var(X_i)} = \sqrt{ 0.25} = 0.5[/tex]
The standard deviation of the total mass of the 4 dots, (T), [tex]\sigma_T[/tex] = 0.5 grams
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