Question

Let X represent the number on the face that lands up when a fair six-sided number cube is tossed. The expected value of X is 3.5, and the standard deviation of X is approximately 1,708. Two fair six-sided number cubes will be tossed and the numbers appearing on the faces that land up will be added.
Which of the following values is closest to the standard deviation of the resulting sum?
(A) 1.708 (B) 1.848 (C) 2.415 (D) 3.416 (E) 5.835

Answer 1

Answer:

c) 2.415

Step-by-step explanation:

Given that X represent the number on the face that lands up when a fair six-sided number cube is tossed.

The expected value of X is 3.5, and the standard deviation of X is approximately 1.708.

When another die is rolled let Y represent the number on the face that lands up when a fair six-sided number cube is tossed.

The expected value of Y is 3.5, and the standard deviation of Y is approximately 1.708.

Also we find that X and Y are independent

Let U = X+Y

Then we have U as the random variable representing the sum shown by two dice

Since X and Y are independent

$Var(x+y) = Var(x) +Var(y)\\= 1.708^2 *2\\= 5.83333$

Std dev for sum

= $\sqrt{5.8333} \\=2.4152$

Hence option C 2.415 is correct

Answer 2

Answer:

2.415

Step-by-step explanation:

Its not as complicated as that other response. You never subtract standard deviations, you always add them, and you don't add them directly, you have to square them to make them variances, add them, and find the square root of it.

In this problem, you do $\sqrt{(1.708^{2}) + (1.708^{2})}$

Put it in the calculator and you get 2.415