Question

PLEASE help me I am studying for the final

The Attila Barbell Company makes bars for weight lifting. The weights of the bars are independent and are normally distributed with a mean of 720 ounces (45 pounds) and a standard deviation of 4 ounces. The bars are shipped 10 in a box to the retailers.The weights of the empty boxes are normally distributed with a mean of 320 ounces and a standard deviation of 8 ounces. The weights of the boxes filled with 10 bars are expected to be normally distributed with a mean of 7,520 ounces and a standard deviation of

Answer 1

Answer:√224 or approximately 14.97 ounces

Explanation: standard deviation of filled boxes=√[(standard deviation of individual bars)^2 x number of bars + (standard deviation of empty boxes)^2]

Then, by plugging in the values given in the problem we get:

standard deviation of filled boxes=√[(4 ounces)^2 x 10 + (8 ounces)^2]

standard deviation of filled boxes=square root [160 + 64] =√224 or 14.97 ounces.

Answer 2

Approximately 8.944 ounces.

Step-by-step explanation:

To find the standard deviation of the weights of the boxes filled with 10 bars, we need to calculate the standard deviation of the sum of two normally distributed variables. In this case, the two variables are the weights of the bars and the weights of the empty boxes.

Since the weights of the bars and the weights of the empty boxes are independent, we can use the formula for the standard deviation of the sum of two independent random variables:

Standard deviation of the sum = sqrt(standard deviation of the first variable^2 + standard deviation of the second variable^2)

In this case, the standard deviation of the weights of the bars is 4 ounces and the standard deviation of the weights of the empty boxes is 8 ounces. Plugging these values into the formula, we get:

Standard deviation of the sum = sqrt(4^2 + 8^2) = sqrt(16 + 64) = sqrt(80) = 8.944

Therefore, the standard deviation of the weights of the boxes filled with 10 bars is approximately 8.944 ounces.