Question

2.6.5 A plant physiologist grew birch seedlings in the green-house and measured the ATP content of their roots. (See Example 1.1.3.) The results (nmol ATP/mg tissue) were as follows for four seedlings that had been handled identically.39 1.45 1.19 1.05 1.07 Calculate the mean and the S

Answer 1

`(a)\ \bar x = 1.19`

`(b)\ \sigma_x = 0.18`

Explanation:

Given

`n = 4`

`x: 1.45\ 1.19\ 1.05\ 1.07`

Solving (a): The mean

This is calculated as:

`\bar x = (\sum x)/(n)`

So, we have:

`\bar x = (1.45 + 1.19 + 1.05 + 1.07)/(4)`

`\bar x = (4.76)/(4)`

`\bar x = 1.19`

Solving (b): The standard deviation

This is calculated as:

`\sigma_x = \sqrt{(\sum(x - \bar x)^2)/(n - 1)}`

So, we have:

`\sigma_x = \sqrt{((1.45 -1.19)^2 + (1.19 -1.19)^2 + (1.05 -1.19)^2 + (1.07 -1.19)^2)/(4 - 1)}`

`\sigma_x = \sqrt{((0.1016)/(3)}`

`\sigma_x = √(0.033867)`

`\sigma_x = 0.18`