Question

Combining random variables
A large bakery uses a machine to dispense frosting onto its cookies. A certain type of cookie gets 4 dots of
frosting on top of each cookie. The dots of frosting have a mean mass of u = 3 grams and a standard deviation
of 0 = 0.25 grams. We can assume that the masses of the dots of frosting are independent from each other.
Let T be the total mass of the 4 dots of frosting on a randomly chosen cookie.
Find the standard deviation of T.
Choose 1 answer:
OT = 0.25 grams
B
OT = 0.5 grams
OT = 0.75 grams
OT 1 gram

Answer 1

Answer:

0.5 grams

Step-by-step explanation:

Answer 2

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; $\underline{\sigma_T = 0.5 \ grams}$

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;

• $\displaystyle Var \left(\sum_{i=1}^{n} X_i \right) = \mathbf{ \sum_{i = 1}^{n}Var(X_i)}$

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;

$\displaystyle \mathbf{Var \left(\sum_{i=1}^{4} X_i \right)} = \sum_{i = 1}^{4}Var(X_i) = Var(X_1) + Var(X_2) +Var(X_3) +Var(X_4)$

Var(X₁) + Var(X₂) + Var(X₃) + Var(X₄) = 0.25² + 0.25² + 0.25² + 0.25² = 0.25

Therefore;

$\displaystyle Var \left(\sum_{i=1}^{4} X_i \right) =\sigma^2_{1 + 2 + 3 +4} = \sum_{i = 1}^{4}Var(X_i) = \mathbf{0.25}$

$\displaystyle \sigma^2_T = \sigma^2_{1 + 2 + 3 +4}$

$\displaystyle Var \left(\sum_{i=1}^{4} X_i \right) =\sigma^2_T = \sum_{i = 1}^{4}Var(X_i) = 0.25$

$\displaystyle Standard \ deviation \ of \ the \ total \ mass = \sigma_T = \sqrt{ \sum_{i = 1}^{4}Var(X_i)} = \sqrt{ 0.25} = 0.5$

The standard deviation of the total mass of the 4 dots, (T), $\sigma_T$ = 0.5 grams

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