A data set lists weights​ (lb) of plastic discarded by households. The highest weight is 5.31 ​lb, the mean of all of the weights is x=2.088 ​lb, and the standard deviation of the weights is s=1.968 lb. a. What is the difference between the weight of 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 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 5.31 lb​ significant?

a) The difference is of 3.222 lbs.

b) 1.64 standard deviations.

c) Z = 1.64

d) Not significant, as the z-score of 1.64 is between -2 and 2.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The mean of all of the weights is x=2.088 ​lb, and the standard deviation of the weights is s=1.968 lb.

This means that

a. What is the difference between the weight of 5.31 lb and the mean of the​ weights?

This is

The difference is of 3.222 lbs.

b. How many standard deviations is that​ [the difference found in part​ (a)]?

This is the z-score. So

1.64 standard deviations.

c. Convert the weight of 5.31 lb to a z score.

Z = 1.64, as found above.

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 5.31 lb​ significant?

Not significant, as the z-score of 1.64 is between -2 and 2.