Question

A homeowner has 5 zucchini plants in her garden. Over the course of the season, the yields (number of zucchinis per plant) are: Plant 1 2 3 4 5 Yield 15 12 17 14 22 Using the information in the table provided, to the nearest tenth, calculate the average yield per plant and the standard deviation. A. average yield per plant: a0 B. standard deviation: a1

Answer 1

Average = 16, standard deviation =

Explanation:

15+12=27

27+(17+14)+22=27+31+22=80

80/5=16

Distance from 22 to 16=6, from 12 to 16=4, so deviation = 5

Answer 2

Average yield per plant is 16 and standard deviation is 3.4058.

Explanation:

Given : A homeowner has 5 zucchini plants in her garden.

Plant 1 2 3 4 5

Yield 15 12 17 14 22

Average :
`\frac{\text {Sum of all yields}}{\text{Total no. of plants}}`

Average :
`(15+12+17+14+22)/(5)`

Average :
`16`

Standard deviation =
`\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}`

=
`\sqrt{((15-16)^2+(12-16)^2+(17-16)^2+(14-16)^2+(22-16)^2)/(5)`

=
`3.4058`

Hence average yield per plant is 16 and standard deviation is 3.4058.