A data set lists weights (lb) of plastic discarded by households. The highest weight is 5.65 lb, the mean of all of the weights is x=2.135 lb, and the standard deviation of the weights is s=2.316 lb. a. What is the difference between the weight of 5.65 lb and the mean of the weights? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the weight of 5.65 lb to a z score. d. If we consider weights that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the weight of 5.65 lb significant?

Explained below.

Step-by-step explanation:

Let the random variable X represent the weights (lb) of plastic discarded by households.

It is provided that the mean weight is, and the standard deviation of the weights is, .

(a)

Compute the difference between the weight of 5.65 lb and the mean of the weights as follows:

Thus, the difference is 3.515 lb.

(b)

Compute the number of standard deviations as follows:

Thus, the number of standard deviation is 1.518.

(c)

Compute the z-score for the weight 5.65 lb as follows:

Thus, the z-score is 1.52.

(d)

The z-score for the weight 5.65 lb is 1.52.

This  z-score lies in the range -2 and 2.

Thus, the weight of 5.65 lb is neither significantly low nor significantly high.

:

step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,