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A data set lists weights​ (lb) of plastic discarded by households. the highest weight is 5.31 5.31 ​lb, the mean of all of the weights is x overbar x equals = 2.257 2.257 ​lb, and the standard deviation of the weights is s equals = 1.655 1.655 lb.

a. what is the difference between the weight of 5.31 5.31 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 5.31 5.31 lb to a z score.


d. if we consider data speeds that convert to z scores between minus −2 and 2 to be neither significantly low nor significantly​ high, is the weight of 5.31 5.31 lb​ significant?

1 Answer

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Answer:

A. = 3.053

B. = 1.85

C. =-1.85

D. since the z sore is less than -2, the highest weight is significantly low

Explanation:

a) Difference between weights

= 5.31 - 2.527

= 3.053

b) Number of standard deviations = = 3.053÷1.655

= 1.85

c) the z score = \frac{\bar{X}-5.26}{s}

= (2.527 - 5.31 )÷1.655

= −3.053 ÷ 1.655

=-1.85

d) since the z sore is less than -2, the highest weight is significantly low

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