Question

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?

Answer 1

Answer:

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, $\bar x=2.135\ \text{lb}$ and the standard deviation of the weights is, $s=2.316\ \text{lb}$.

(a)

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

$d=5.65 - \bar x\\\\d=5.65-2.135\\\\d=3.515$

Thus, the difference is 3.515 lb.

(b)

Compute the number of standard deviations as follows:

$\text{Number of Standard Deviation}=\frac{d}{s}=\frac{3.515}{2.316}=1.518$

Thus, the number of standard deviation is 1.518.

(c)

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

$z=\frac{a-\bar x}{s}=\frac{5.65-2.135}{2.316}=1.517703\approx 1.52$

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.