Respuesta :
(1) Frames covered with shade cloth mean = 1.71 and standard deviation = 0.401
(2) control pvc frames only mean = 19.26 and standard deviation = 0.536
(3) warmer (pvc frames covered with plastic) mean = 2.08 and standard deviation = 0.774.
(1) Using the cooler mean and SD:
The calculations are shown in the table below:
[tex]\begin{array}{lll} & \mathrm{X} & (\mathrm{X} \text {-mean })^{2} \\& 1.59 & 0.0144 \\& 1.43 & 0.0784 \\& 1.88 & 0.0289 \\& 1.26 & 0.2025 \\& 1.91 & 0.04 \\& 1.86 & 0.0225 \\& 1.9 & 0.0361 \\& 1.57 & 0.0196 \\& 1.79 & 0.0064 \\& 1.72 & 0.0001 \\& 2.41 & 0.49 \\& 2.34 & 0.3969 \\& 0.83 & 0.7744 \\& 1.34 & 0.1369 \\& 1.76 & 0.0025 \\\text { Total } & 25.59 & 2.2496\end{array}[/tex]
Sample size
[tex]$n=15$[/tex]
Sample sum:
[tex]$\sum x=25.59$[/tex]
Sample mean:
[tex]$\begin{aligned}\bar{x}=\frac{\sum x}{n} &=25.59 / 15 \\&=1.71\end{aligned}$[/tex]
Sample standard deviation:
[tex]\begin{aligned}s=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} &=\sqrt(2.2496 /(15-1)) \\&=0.401\end{aligned}[/tex]
(2) Using the control mean and SD:
The calculations are shown in the table below:
[tex]\begin{array}{lll} & \mathrm{X} & (\mathrm{X}-\text { mean })^{2} \\& 1.92 & 0.0289 \\& 2 & 0.0625 \\& 2.19 & 0.1936 \\& 1.12 & 0.3969 \\& 1.78 & 0.0009 \\& 1.84 & 0.0081 \\& 2.45 & 0.49 \\& 2.03 & 0.0784 \\& 1.52 & 0.0529 \\& 0.51 & 1.5376 \\& 1.9 & 0.0225 \\\text { Total } & 19.26 & 2.8723\end{array}[/tex]
Sample size:
[tex]$n=11$[/tex]
Sample sum:
[tex]$\sum x=19.26$[/tex]
Sample mean:
[tex]\begin{aligned}s=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} &=\sqrt(2.8723 /(11-1)) \\&=0.536\end{aligned}[/tex]
(3) Assuming a warmer mean and SD:
The calculations are shown in the table below:
[tex]$\begin{array}{lll} & \mathrm{X} & (\text { X-mean })^{2} \\ & 2.57 & 0.3969 \\ & 2.6 & 0.4356 \\ & 1.93 & 0.0001 \\ & 1.58 & 0.1296 \\ & 2.3 & 0.1296 \\ & 0.84 & 1.21 \\ & 2.65 & 0.5041 \\ & 0.14 & 3.24 \\ & 2.74 & 0.64 \\ & 2.53 & 0.3481 \\ & 2.13 & 0.0361 \\ & 2.86 & 0.8464 \\ & 2.31 & 0.1369 \\ & 1.91 & 0.0009 \\ \text { Total } & 29.09 & 8.0543\end{array}$[/tex]
Sample size:
[tex]$n=14$[/tex]
Sample sum:
[tex]$\sum x=29.09$[/tex]
Sample mean:
[tex]$\begin{aligned}\bar{x}=\frac{\sum x}{n} &=29.09 / 14 \\&=2.08\end{aligned}$[/tex]
Sample standard deviation:
[tex]s=\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}} &=\sqrt(7.7883 /(14-1)) \\&[/tex]
[tex]=0.774[/tex]
Hence, these are the different samples of mean and standard deviation.
What does "standard deviation" mean?
The term "standard deviation" (or "") refers to a measurement of the data's dispersion from the mean. A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed.
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