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A data set lists weights​ (lb) of plastic discarded by households. The highest weight is 5.29 ​lb, the mean of all of the weights is x=1.995 ​lb, and the standard deviation of the weights is s=1.305 lb. a. What is the difference between the weight of 5.29 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 5.29 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 5.29 lb​ significant?

User Zhengchun
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1 Answer

1 vote

Answer:

Follows are the solution to this question:

Explanation:

In point a:

The weight difference between 5.29 lb and the weighted mean:


=5.29-1.995 \\\\=3.295

In point b:

The number of standard deviation:


=(3.295)/(1.305) \\\\ =2.52

In point c:


z score= (x- \mu)/(\sigma)\\\\


=( 1.995- 5.29)/(1.305)\\\\=(-3.295)/(1.305)\\\\= - 2.52

In point d:

No, since (-2, 2) included with the Zero is the z score and for weight is 5.29.

User Geethanga
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