Answer:
(a) The difference between the highest weight and mean weight is 3.051 lb.
(b) The number of standard deviations is 2.77.
(c) The <em>z</em>-score of 5.26 is 2.77.
(d) The weight of 5.26 lb is significantly high.
Step-by-step explanation:
The random variable <em>X</em> is defined as the weights (lb) of plastic discarded by households.
The highest weight is, ![X_{max.}=5.26\ lb](https://tex.z-dn.net/?f=X_%7Bmax.%7D%3D5.26%5C%20lb)
The mean weight is,
.
The standard deviation of the weight is,
.
(a)
Compute the difference between the highest weight and mean weight as follows:
![X_{max.}-\bar x=5.26-2.209=3.051](https://tex.z-dn.net/?f=X_%7Bmax.%7D-%5Cbar%20x%3D5.26-2.209%3D3.051)
Thus, the difference between the highest weight and mean weight is 3.051 lb.
(b)
Compute the number of standard deviations the mean is from the maximum value as follows:
![Number\ of\ standard deviation=\frac{X_{max.}-\bar x}{s}=\frac{3.051}{1.0101}=2.77](https://tex.z-dn.net/?f=Number%5C%20of%5C%20standard%20deviation%3D%5Cfrac%7BX_%7Bmax.%7D-%5Cbar%20x%7D%7Bs%7D%3D%5Cfrac%7B3.051%7D%7B1.0101%7D%3D2.77)
Thus, the number of standard deviations is 2.77.
(c)
The formula of <em>z</em>-score is:
![z=\frac{X-\bar x}{s}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BX-%5Cbar%20x%7D%7Bs%7D)
Compute the <em>z</em>-score for <em>X</em> = 5.26 as follows:
![z=\frac{X-\bar x}{s}=\frac{5.26-2.209}{1.101}=2.77](https://tex.z-dn.net/?f=z%3D%5Cfrac%7BX-%5Cbar%20x%7D%7Bs%7D%3D%5Cfrac%7B5.26-2.209%7D%7B1.101%7D%3D2.77)
Thus, the <em>z</em>-score of 5.26 is 2.77.
(d)
The <em>z</em>-scores between -2 and 2 are considered as neither significantly low nor significantly high.
The <em>z</em>-score for <em>X</em> = 5.26 is 2.77.
The value of <em>z</em> > 2.
Thus, the weight of 5.26 lb is significantly high.