A data set lists weights (lb) of plastic discarded by households. The highest weight is

5.265.26

lb, the mean of all of the weights is

x overbarxequals=2.2092.209

lb, and the standard deviation of the weights is

sequals=1.1011.101

lb. a. What is the difference between the weight of

5.265.26

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

5.265.26

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

5.265.26

lb significant?

a. The difference is

nothing

lb.

(Type an integer or a decimal. Do not round.)

b. The difference is

nothing

standard deviations.

(Round to two decimal places as needed.)

c. The z score is

zequals=nothing.

(Round to two decimal places as needed.)

d. The highest weight is

(a) The difference between the highest weight and mean weight is 3.051 lb.

(b) The number of standard deviations is 2.77.

(c) The z-score of 5.26 is 2.77.

(d) The weight of 5.26 lb is significantly high.

Step-by-step explanation:

The random variable X is defined as the weights (lb) of plastic discarded by households.

The highest weight is,

The mean weight is, .

The standard deviation of the weight is, .

(a)

Compute the difference between the highest weight and mean weight as follows:

Thus, the difference between the highest weight and mean weight is 3.051 lb.

(b)

Compute the number of standard deviations the mean is from the maximum  value as follows:

Thus, the number of standard deviations is 2.77.

(c)

The formula of z-score is:

Compute the z-score for X = 5.26 as follows:

Thus, the z-score of 5.26 is 2.77.

(d)

The z-scores between -2 and 2 are considered as neither significantly low nor significantly high.

The z-score for X = 5.26  is 2.77.

The value of z > 2.

Thus, the weight of 5.26 lb is significantly high.

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step-by-step explanation:

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