Question

a data set lists weights (lb) of plastic discarded by households. the highest weight is 5.13 lb, the mean of all of the weights is \\( \bar { x } = 2.249 \\) lb, and the standard deviation of the weights is \\( s = 1.762 \\) lb.
a. what is the difference between the weight of 5.13 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 5.13 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 5.13 lb significant?
a. the difference is 2.881 lb
(type an integer or a decimal. do not round.)
b. the difference is 1.64 standard deviations
(round to two decimal places as needed)
c. the z score is z = \\( \square \\)
(round to two decimal places as needed)

Explanation:

Step1: Calculate the difference between the weight and the mean

The formula for the difference is \(x_{value}-\bar{x}\).
Here, \(x_{value} = 5.13\) lb and \(\bar{x}=2.249\) lb.
So, \(5.13 - 2.249=2.881\)

Step2: Calculate the number of standard deviations

The formula is \(\frac{x_{value}-\bar{x}}{s}\).
We know \(x_{value}-\bar{x}=2.881\) lb and \(s = 1.76\) lb.
So, \(\frac{2.881}{1.76}\approx1.64\)

Step3: Calculate the z - score

The z - score formula is \(z=\frac{x-\mu}{\sigma}\) (same as the formula in step 2).
\(z=\frac{5.13 - 2.249}{1.76}=\frac{2.881}{1.76}\approx1.64\)

Answer:

a. \(2.881\)
b. \(1.64\)
c. \(1.64\)
d. Since \(z = 1.64\) and \(- 2<1.64<2\), the weight of \(5.13\) lb is not significant.