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?

A. = 3.053

B. = 1.85

C. =-1.85

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

Step-by-step 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

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

i think it is c, not positive though

step-by-step explanation: