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: B. Standard deviation:.

Answer 1

The mean and standard deviation is 16 and 12.6.

Standard deviation

It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.

$\rm \sigma = \sqrt{\dfrac{\Sigma(x_i - \mu)^2}{N}}\\\\Where \\\sigma = Standard\ deviation\\ \mu = mean\ population\\x_i = each\ value\ of\ population\\N = size\ of \ the \ population$

Mean

Mean is simply defined as the average of the given set of numbers. The mean is said to be an arithmetic mean. It is the ratio of the sum of the observation to the total number of observations.

$\rm Mean = \dfrac{sum\ of\ the\ observation}{number\ of\ observation}$

Given

A homeowner has 5 zucchini plants in her garden.

Plant 1, 2, 3, 4, 5 and their Yield 15, 12, 17, 14, 22.

To find

The mean and standard deviation.

How to find it?

1.  Mean will be

$\rm Mean = \dfrac{15+12+17+14+22}{5}\\\\Mean = \dfrac{80}{5}\\\\Mean = 16$

2.  Standard deviation will be

$\rm \sigma = \sqrt{\dfrac{\Sigma(x_i - \mu)^2}{N}}\\\\\sigma = \sqrt{\dfrac{63}{5}}\\\\\sigma = 12.6$

The mean and standard deviation is 16 and 12.6.

More about the standard deviation link is given below.

https://brainly.com/question/12402189