Question: User: Number of brown lentils per quadrat\nGroup Number\t0\t1\t2\t3\t4\t5\t6\t7\t8\t9\nGroup 1\t0\t0\t0\t1\t7\t6\t0\t4\t1\t0\nGroup 2\t0\t0\t0\t1\t4\t6\t2\t2\t0\t0\nGroup 3\t0\t0\t0\t0\t7\t5\t5\t0\t2\t0\nGroup 4\t0\t0\t0\t0\t6\t6\t3\t1\t1\t0\nGroup 5\t0\t0\t0\t1\t8\t6\t1\t0\t1\t0\nGroup 6\t0\t0\t0\t1\t5\t6\t3\t2\t1\t0\nGroup 7\t0\t0\t0\t0\t7\t4\t2\t0\t1\t0\nGroup 8\t0\t0\t0\t0\t7\t6\t6\t0\t1\t0\nTotal\t0\t0\t0\t4\t51\t45\t22\t9\t8\t0sing the data shown in the Table attached to this Tests and Quizzes page, calculate the variance (s2) around your estimate of the average number of brown lentils per quadrat observed by Group 6. Your answer should be given to TWO decimal places.
1. **State the problem:** Calculate the variance $s^2$ of the number of brown lentils per quadrat observed by Group 6.
2. **Data for Group 6:** The counts for lentils per quadrat are given by the frequencies for each number of lentils (0 to 9):
$$\text{Values } x = 0,1,2,3,4,5,6,7,8,9$$
$$\text{Frequencies } f = 0,0,0,1,5,6,3,2,1,0$$
3. **Calculate the mean $\bar{x}$:**
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{0\times0 + 0\times1 + 0\times2 + 1\times3 + 5\times4 + 6\times5 + 3\times6 + 2\times7 + 1\times8 + 0\times9}{0+0+0+1+5+6+3+2+1+0}$$
Calculate numerator:
$$= 0 + 0 + 0 + 3 + 20 + 30 + 18 + 14 + 8 + 0 = 93$$
Calculate denominator:
$$= 0 + 0 + 0 + 1 + 5 + 6 + 3 + 2 + 1 + 0 = 18$$
So,
$$\bar{x} = \frac{93}{18} = 5.1667$$
4. **Calculate variance $s^2$ formula:**
$$s^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i - 1}$$
5. **Calculate each squared deviation and multiply by frequency:**
\begin{align*}
f_i (x_i - \bar{x})^2 &= f_i (x_i^2 - 2x_i\bar{x} + \bar{x}^2) \\
\end{align*}
Calculate each term:
- For $x=3$, $f=1$: $1 \times (3 - 5.1667)^2 = 1 \times ( -2.1667)^2 = 4.6944$
- For $x=4$, $f=5$: $5 \times (4 - 5.1667)^2 = 5 \times ( -1.1667)^2 = 5 \times 1.3611 = 6.8055$
- For $x=5$, $f=6$: $6 \times (5 - 5.1667)^2 = 6 \times ( -0.1667)^2 = 6 \times 0.0278 = 0.1667$
- For $x=6$, $f=3$: $3 \times (6 - 5.1667)^2 = 3 \times 0.8333^2 = 3 \times 0.6944 = 2.0833$
- For $x=7$, $f=2$: $2 \times (7 - 5.1667)^2 = 2 \times 1.8333^2 = 2 \times 3.3611 = 6.7222$
- For $x=8$, $f=1$: $1 \times (8 - 5.1667)^2 = 1 \times 2.8333^2 = 8.0278$
Sum these values:
$$4.6944 + 6.8055 + 0.1667 + 2.0833 + 6.7222 + 8.0278 = 28.5$$
6. **Calculate variance:**
$$s^2 = \frac{28.5}{18 - 1} = \frac{28.5}{17} = 1.6765$$
Rounded to two decimal places:
$$s^2 = 1.68$$
**Final answer:** The variance of the number of brown lentils per quadrat observed by Group 6 is **1.68**.