91影视

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.

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

$z=\frac{x-\stackrel{炉}{x}}{s}$

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

a.

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

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

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:

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

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:

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

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.