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

Answer:

$(a)\ \bar x = 1.19$

$(b)\ \sigma_x = 0.18$

Step-by-step explanation:

Given

$n = 4$

$x: 1.45\ 1.19\ 1.05\ 1.07$

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{1.45 + 1.19 + 1.05 + 1.07}{4}$

$\bar x = \frac{4.76}{4}$

$\bar x = 1.19$

Solving (b): The standard deviation

This is calculated as:

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

So, we have:

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

$\sigma_x = \sqrt{\frac{(0.1016}{3}}$

$\sigma_x = \sqrt{0.033867}$

$\sigma_x = 0.18$