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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

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MrRoyal

Answer:

[tex](a)\ \bar x = 1.19[/tex]

[tex](b)\ \sigma_x = 0.18[/tex]

Step-by-step explanation:

Given

[tex]n = 4[/tex]

[tex]x: 1.45\ 1.19\ 1.05\ 1.07[/tex]

Solving (a): The mean

This is calculated as:

[tex]\bar x = \frac{\sum x}{n}[/tex]

So, we have:

[tex]\bar x = \frac{1.45 + 1.19 + 1.05 + 1.07}{4}[/tex]

[tex]\bar x = \frac{4.76}{4}[/tex]

[tex]\bar x = 1.19[/tex]

Solving (b): The standard deviation

This is calculated as:

[tex]\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n - 1}}[/tex]

So, we have:

[tex]\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}}[/tex]

[tex]\sigma_x = \sqrt{\frac{(0.1016}{3}}[/tex]

[tex]\sigma_x = \sqrt{0.033867}[/tex]

[tex]\sigma_x = 0.18[/tex]

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