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a data set lists weights (lb) of plastic discarded by households. the h…

Question

a data set lists weights (lb) of plastic discarded by households. the highest weight is 5.51 lb, the mean of all of the weights is ( overline{x}=2.174 ) lb, and the standard deviation of the weights is ( s = 1.986 ) lb.
a. what is the difference between the weight of 5.51 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 5.51 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 5.51 lb significant?
a. the difference is 3.336 lb.
(type an integer or a decimal. do not round.)
b. the difference is ( square ) standard deviations.
(round to two decimal places as needed )

Explanation:

Step1: Calculate the number of standard deviations

The formula to find the number of standard deviations a value is from the mean is \(n=\frac{\text{Value}-\text{Mean}}{\text{Standard Deviation}}\).
We know the value \(x = 5.51\), the mean \(\bar{x}=2.174\), and the standard deviation \(s = 1.986\).
The difference from part (a) is \(x-\bar{x}=5.51 - 2.174=3.336\).
To find how many standard deviations this difference is, we use \(n=\frac{x-\bar{x}}{s}\).
Substitute the values: \(n=\frac{3.336}{1.986}\).

Step2: Perform the division

\(n=\frac{3.336}{1.986}\approx1.68\) (rounded to two decimal places).

Answer:

1.68