find the range and the standard deviation for the three samples below.
sample a: 41, 43, 45, 47, 49, 51, 53
sample b: 41, 42, 43, 47, 51, 52, 53
sample c: 41, 41, 41, 47, 53, 53, 53

Question

Accepted Answer

# Explanation:
## Step1: Calculate the range
The range is calculated as $Range = Max - Min$.
- For Sample A: $Max = 53$, $Min = 41$, so $Range_A=53 - 41=12$.
- For Sample B: $Max = 53$, $Min = 41$, so $Range_B=53 - 41 = 12$.
- For Sample C: $Max = 53$, $Min = 41$, so $Range_C=53 - 41=12$.

## Step2: Calculate the mean ($\bar{x}$)
The formula for the mean is $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$.
- For Sample A: $n = 6$, $\sum x=(41 + 43+45 + 47+49+51+53)$
$$
\begin{align*}
\bar{x}_A&=\frac{41+43 + 45+47+49+51+53}{6}\
&=\frac{(41 + 53)+(43 + 51)+(45 + 49)+47}{6}\
&=\frac{94+94+94 + 47}{6}\
&=\frac{329}{6}\approx48.17
\end{align*}
$$
- For Sample B: $n = 6$, $\sum x=(41+42 + 43+47+51+52+53)$
$$
\begin{align*}
\bar{x}_B&=\frac{41+42+43+47+51+52+53}{6}\
&=\frac{(41+53)+(42 + 52)+(43+51)+47}{6}\
&=\frac{94+94+94+47}{6}\
&=\frac{329}{6}\approx48.17
\end{align*}
$$
- For Sample C: $n = 6$, $\sum x=(41+41 + 41+47+53+53+53)$
$$
\begin{align*}
\bar{x}_C&=\frac{41\times3+47+53\times3}{6}\
&=\frac{123+47 + 159}{6}\
&=\frac{329}{6}\approx48.17
\end{align*}
$$

## Step3: Calculate the standard deviation ($s$)
The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$
- For Sample A:
$$
\begin{align*}
\sum(x_i-\bar{x}_A)^2&=(41 - 48.17)^2+(43-48.17)^2+(45 - 48.17)^2+(47-48.17)^2+(49-48.17)^2+(51 - 48.17)^2+(53-48.17)^2\
&=(-7.17)^2+(-5.17)^2+(-3.17)^2+(-1.17)^2+(0.83)^2+(2.83)^2+(4.83)^2\
&=51.4089+26.7289 + 10.0489+1.3689+0.6889+8.0089+23.3289\
&=121.5933
\end{align*}
$$
$s_A=\sqrt{\frac{121.5933}{5}}\approx\sqrt{24.3187}\approx4.93$

- For Sample B:
$$
\begin{align*}
\sum(x_i-\bar{x}_B)^2&=(41 - 48.17)^2+(42-48.17)^2+(43 - 48.17)^2+(47-48.17)^2+(51-48.17)^2+(52 - 48.17)^2+(53-48.17)^2\
&=(-7.17)^2+(-6.17)^2+(-5.17)^2+(-1.17)^2+(2.83)^2+(3.83)^2+(4.83)^2\
&=51.4089+38.0689+26.7289+1.3689+8.0089+14.6689+23.3289\
&=163.5823
\end{align*}
$$
$s_B=\sqrt{\frac{163.5823}{5}}\approx\sqrt{32.7165}\approx5.72$

- For Sample C:
$$
\begin{align*}
\sum(x_i-\bar{x}_C)^2&=(41 - 48.17)^2+(41-48.17)^2+(41 - 48.17)^2+(47-48.17)^2+(53-48.17)^2+(53 - 48.17)^2+(53-48.17)^2\
&=3\times(-7.17)^2+(-1.17)^2+3\times(4.83)^2\
&=3\times51.4089+1.3689+3\times23.3289\
&=154.2267+1.3689+69.9867\
&=225.5823
\end{align*}
$$
$s_C=\sqrt{\frac{225.5823}{5}}\approx\sqrt{45.1165}\approx6.72$

# Answer:
- **Range**: For Sample A, Sample B and Sample C, the range is $12$.
- **Standard Deviation**: $s_A\approx4.93$, $s_B\approx5.72$, $s_C\approx6.72$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the range

Step2: Calculate the mean (\(\bar{x}\))

Step3: Calculate the standard deviation (\(s\))

Answer: