Question

In Exercises 5–8, express all z scores with two decimal places.

Plastic Waste Data Set 31 “Garbage Weight” in Appendix B lists weights (lb) of plastic discarded by households. The highest weight is 5.28 lb, the mean of all of the weights is x = 1.911 lb, and the standard deviation of the weights is s = 1.065 lb.

a. What is the difference between the weight of 5.28 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 5.28 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 5.28 lb significant?

Short answer

a. The difference between the value of weight equal to 5.28 lb and the mean weight is equal to 3.369 lb.

b. The difference between the value of weight equal to 5.28 lb and the mean weight is 3.16 standard deviations.

c. 3.16 is the z-score value for the given weight of 5.28 lb.

d. The weight equal to 5.28 lb is significantly high.

Step-by-step solution

Given information

A sample is provided that shows the weight of plastic discarded by 62 households.

The value of the mean weight is 1.911 lb, and the value of the standard deviation of the weights is 1.065 lb.

Z-score

The value of thez-scoreis obtained for a given data point (x) by using the sample mean $\left(\overline{x}\right)$ and the sample standard deviation (s) as shown in the given formula:

$z=\frac{x-\overline{x}}{s}$

It has no units and estimates the position of the data point above or below the mean.

Calculation

a.

The difference between the given value of weight (5.28 lb) and the mean weight (1.911 lb) is calculated below:

$$\begin{gathered} x-\overline{x}=5.28-1.911 \\ =3.369 \end{gathered}$$

Thus, the difference between the given value of weight (5.28 lb) and the mean weight (1.911 lb) is3.369 lb.

b.

The z-score for the given value of weight (5.28 lb) is calculated below:

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{5.28-1.911}{1.065} \\ =3.16 \end{gathered}$$

Therefore, the difference between the given weight and the mean weight is3.16 standard deviations.

c.

The z-score for the given value of weight (5.28 lb) is calculated below:

$$\begin{gathered} z=\frac{x-\overline{x}}{s} \\ =\frac{5.28-1.911}{1.065} \\ =3.16 \end{gathered}$$

Here, the calculated value of the z-score is equal to 3.16.

d.

Since the z-score value is above 2, it can be said that the value of weight equal to 5.28 lb is significantly high.