Answer:
So, the sample mean is 9.99.
So, the standard deviation is 0.32.
Step-by-step explanation:
From Exercise we have the next numbers: 9.7 9.9 10.3 10.1 10.5 9.4 9.9 10.1 9.7 10.3.
So, N=10, because we have 10 numbers.
We calculate the sample mean:
[tex]\overline{x}=\frac{9.7+ 9.9+ 10.3+ 10.1+ 10.5+ 9.4+ 9.9+ 10.1+ 9.7+ 10.3}{10}\\\\\overline{x}=\frac{99.9}{10}\\\\\overline{x}=9.99[/tex]
So, the sample mean is 9.99.
We use the formula for standard deviation:
[tex]\sigma=\sqrt{\frac{1}{N}\sum_{i=1}^N(x_i-\overline{x})^2}[/tex]
Now, we calculate the sum
[tex]\sum_{i=1}^{10}(x_i-9.99)^2=(9.7-9.99)^2+(9.9-9.99)^2+(10.3-9.99)^2+(10.1-9.99)^2+(10.5-9.99)^2+(9.4-9.99)^2+(9.9-9.99)^2+(10.1-9.99)^2+(9.7-9.99)^2+(10.3-9.99)^2\\\\\sum_{i=1}^{10}(x_i-9.99)^2=1.009[/tex]
Therefore, we get
[tex]\sigma=\sqrt{\frac{1}{N}\sum_{i=1}^N(x_i-\overline{x})^2}\\\\\sigma=\sqrt{\frac{1}{10}\cdot 1.009}\\\\\sigma=0.32[/tex]
So, the standard deviation is 0.32.
