Question

a data set lists weights (lb) of plastic discarded by households. the highest weight is 5.34 lb, the mean of all of the weights is $\bar{x}=2.009$ lb, and the standard deviation of the weights is $s = 1.754$ lb. a. what is the difference between the weight of 5.34 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 5.34 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 5.34 lb significant? a. the difference is 3.331 lb. (type an integer or a decimal. do not round.) b. the difference is standard deviations. (round to two decimal places as needed.)

Explanation:

Step1: Calculate number of standard - deviations

To find how many standard deviations the difference is, we divide the difference from part (a) by the standard deviation. The difference from part (a) is $5.34 - 2.009=3.331$ lb and the standard deviation $s = 1.754$ lb. The formula is $\frac{5.34 - 2.009}{s}$.
$\frac{3.331}{1.754}\approx1.8985$

Step2: Convert to z - score

The z - score formula is $z=\frac{x-\bar{x}}{s}$, where $x = 5.34$, $\bar{x}=2.009$ and $s = 1.754$. This is the same calculation as in part (b), so $z=\frac{5.34 - 2.009}{1.754}\approx1.90$ (rounded to two decimal places).

Step3: Determine significance

We are given that z - scores between - 2 and 2 are neither significantly low nor significantly high. Since $z\approx1.90$ and $- 2<1.90<2$, the weight of 5.34 lb is not significant.

Answer:

b. 1.90
c. 1.90
d. No