(a)

$$\begin{align*} \sum fx^2 & = 1 \times 3^2 + 1 \times 4^2 + 1 \times 5^2 + 1 \times 7^2 + 1 \times 8^2 + 1 \times 10^2 + 1 \times 13^2 \\ & = 432 \\ \\ \sum f & = 7 \\ \\ \sum fx & = 1 \times 3 + 1 \times 4 + 1 \times 5 + 1 \times 7 + 1 \times 8 + 1 \times 10 + 1 \times 13 \\ & = 50 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( {\sum fx \over \sum f} \right)^2 } \\ & = \sqrt{ {432 \over 7} - \left(50 \over 7\right)^2 } \\ & = 3.27014 \\ & \approx 3.27 \end{align*}$$

(b)

$$\begin{align*} 19, 20, 22, & \phantom{.} 23, 24, 25, 27, 28, 30, 32 \\ \\ \sum fx^2 & = 1 \times 19^2 + 1 \times 20^2 + 1 \times 22^2 + 1 \times 23^2 + 1 \times 24^2 + 1 \times 25^2 + 1 \times 27^2 + 1 \times 28^2 + 1 \times 30^2 + 1 \times 32^2 \\ & = 6412 \\ \\ \sum f & = 10 \\ \\ \sum fx & = 1 \times 19 + 1 \times 20 + 1 \times 22 + 1 \times 23 + 1 \times 24 + 1 \times 25 + 1 \times 27 + 1 \times 28 + 1 \times 30 + 1 \times 32 \\ & = 250 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( {\sum fx \over \sum f} \right)^2 } \\ & = \sqrt{ {6412 \over 10} - \left(250 \over 10\right)^2 } \\ & = 4.02492 \\ & \approx 4.02 \end{align*}$$

(c)

$$\begin{align*} -5, & \phantom{.} -4, -2, 0, 1, 4 \\ \\ \sum fx^2 & = 1 \times (-5)^2 + 1 \times (-4)^2 + 1 \times (-2)^2 + 1 \times 0^2 + 1 \times 1^2 + 1 \times 4^2 \\ & = 62 \\ \\ \sum f & = 6 \\ \\ \sum fx & = 1 \times (-5) + 1 \times (-4) + 1 \times (-2) + 1 \times 0 + 1 \times 1 + 1 \times 4 \\ & = -6 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( {\sum fx \over \sum f} \right)^2 } \\ & = \sqrt{ {62 \over 6} - \left(-6 \over 6\right)^2 } \\ & = 3.05505 \\ & \approx 3.06 \end{align*}$$

(a) The table in your calculator should look like:

$$\begin{align*} \text{From the calculator, standard deviation} & = 11.06044 \\ & \approx 11.1 \end{align*}$$

(b) The table in your calculator should look like:

$$\begin{align*} \text{From the calculator, standard deviation} & = 9.35414 \\ & \approx 9.35 \end{align*}$$

(c) The table in your calculator should look like:

$$\begin{align*} \text{From the calculator, standard deviation} & = 11.94278 \\ & \approx 11.9 \end{align*}$$

Note: Marks is the data (x) while number of students is the frequency (f)

$$\begin{align*} \sum fx^2 & = 5 \times 2^2 + 7 \times 3^2 + 6 \times 4^2 + 4 \times 5^2 + 9 \times 6^2 + 3 \times 7^2 + 6 \times 8^2 \\ & = 1134 \\ \\ \sum f & = 5 + 7 + 6 + 4 + 9 + 3 + 6 \\ & = 40 \\ \\ \sum fx & = 5 \times 2 + 7 \times 3 + 6 \times 4 + 4 \times 5 + 9 \times 6 + 3 \times 7 + 6 \times 8 \\ & = 198 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f \right)^2 } \\ & = \sqrt{ {1134 \over 40} - \left(198 \over 40\right)^2 } \\ & = 1.9615 \\ & \approx 1.96 \end{align*}$$

Note: Number of goals per match is the data (x) while number of matches is the frequency (f)

$$\begin{align*} \sum fx^2 & = 10 \times 0^2 + 8 \times 1^2 + 7 \times 2^2 + 6 \times 3^2 + 2 \times 4^2 + 3 \times 5^2 + 1 \times 6^2 \\ & = 233 \\ \\ \sum f & = 10 + 8 + 7 + 6 + 2 + 3 + 1 \\ & = 37 \\ \\ \sum fx & = 10 \times 0 + 8 \times 1 + 7 \times 2 + 6 \times 3 + 2 \times 4 + 3 \times 5 + 1 \times 6 \\ & = 69 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f \right)^2 } \\ & = \sqrt{ {233 \over 37} - \left(69 \over 37\right)^2 } \\ & = 1.6781 \\ & \approx 1.68 \end{align*}$$

We need to use the mid-value for x: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5

$$\begin{align*} \sum fx^2 & = 4 \times 2.5^2 + 12 \times 7.5^2 + 20 \times 12.5^2 + 24 \times 17.5^2 + 16 \times 22.5^2 + 4 \times 27.5^2 \\ & = 22300 \\ \\ \sum f & = 4 + 12 + 20 + 24 + 16 + 4 \\ & = 80 \\ \\ \sum fx & = 4 \times 2.5 + 12 \times 7.5 + 20 \times 12.5 + 24 \times 17.5 + 16 \times 22.5 + 4 \times 27.5 \\ & = 1240 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f \right)^2 } \\ & = \sqrt{ {22300 \over 80} - \left(1240 \over 80\right)^2 } \\ & = 6.2048 \\ & \approx 6.20 \end{align*}$$

We need to use the mid-value for Salary: 210, 230, 250, 270, 290

$$\begin{align*} \sum fx^2 & = 8 \times 210^2 + 23 \times 230^2 + 16 \times 250^2 + 3 \times 270^2 + 10 \times 290^2 \\ & = 3,629,200 \\ \\ \sum f & = 8 + 23 + 16 + 3 + 10 \\ & = 60 \\ \\ \sum fx & = 8 \times 210 + 23 \times 230 + 16 \times 250 + 3 \times 270 + 10 \times 290 \\ & = 14680 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f \right)^2 } \\ & = \sqrt{ {3,629,200 \over 60} - \left(14680 \over 60\right)^2 } \\ & = 24.99777 \\ & \approx 25.0 \end{align*}$$

(a) We need to use the mid-value for x: 35, 45, 55, 65, 75

$$\begin{align*} \sum fx^2 & = 16 \times 35^2 + 25 \times 45^2 + 35 \times 55^2 + 14 \times 65^2 + 10 \times 75^2 \\ & = 291,500 \\ \\ \sum f & = 16 + 25 +35 +14 + 10 \\ & = 100 \\ \\ \sum fx & = 16 \times 35 + 25 \times 45 + 35 \times 55 + 14 \times 65 + 10 \times 75 \\ & = 5270 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fx^2 \over \sum f} - \left( \sum fx \over \sum f \right)^2 } \\ & = \sqrt{ {291,500 \over 100} - \left(5270 \over 100\right)^2 } \\ & = 11.7349 \\ & \approx 11.7 \end{align*}$$

(b) We need to use the mid-value for y: 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5

$$\begin{align*} \sum fy^2 & = 4 \times 72.5^2 + 11 \times 77.5^2 + 15 \times 82.5^2 + 24 \times 87.5^2 + 18 \times 92.5^2 + 9 \times 97.5^2 + 3 \times 102.5^2 \\ & = 644,025 \\ \\ \sum f & = 4 + 11 + 15 + 24 + 18 + 9 + 3 \\ & = 84 \\ \\ \sum fy & = 4 \times 72.5 + 11 \times 77.5 + 15 \times 82.5 + 24 \times 87.5 + 18 \times 92.5 + 9 \times 97.5 + 3 \times 102.5 \\ & = 7330 \\ \\ \text{Standard deviation} & = \sqrt{ {\sum fy^2 \over \sum f} - \left( \sum fy \over \sum f \right)^2 } \\ & = \sqrt{ {644,025 \over 84} - \left(7330 \over 84\right)^2 } \\ & = 7.2335 \\ & \approx 7.23 \end{align*}$$

(i)

$$\begin{align*} \text{For Class } A, \sum fx & = 1 \times 4 + 2 \times 6 + 1 \times 7 + 1 \times 8 + 1 \times 10 + 1 \times 11 + 1 \times 12 \\ & = 64 \\ \\ \sum f & = 8 \\ \\ \sum fx^2 & = 1 \times 4^2 + 2 \times 6^2 + 1 \times 7^2 + 1 \times 8^2 + 1 \times 10^2 + 1 \times 11^2 + 1 \times 12^2 \\ & = 566 \\ \\ \text{Mean (Class } A) & = {\sum fx \over \sum f} \\ & = {64 \over 8} \\ & = 8 \\ \\ \text{S.D. (Class } A) & = \sqrt{ {\sum fx^2 \over \sum f} - ( \text{Mean} )^2 } \\ & = \sqrt{{566 \over 8} - (8)^2} \\ & = 2.5980 \\ & \approx 2.60 \\ \\ \\ \text{For Class } B, \sum fx & = 1 \times 0 + 2 \times 1 + 1 \times 2 + 1 \times 3 + 1 \times 14 + 1 \times 17 + 1 \times 25 \\ & = 63 \\ \\ \sum f & = 8 \\ \\ \sum fx^2 & = 1 \times 0^2 + 2 \times 1^2 + 1 \times 2^2 + 1 \times 3^2 + 1 \times 14^2 + 1 \times 17^2 + 1 \times 25^2 \\ & = 1125 \\ \\ \text{Mean (Class } B) & = {63 \over 8} \\ & = 7.875 \\ \\ \text{S.D. (Class } B) & = \sqrt{ {\sum fx^2 \over \sum f} - ( \text{Mean} )^2 } \\ & = \sqrt{{1125 \over 8} - (7.875)^2} \\ & = 8.8661 \\ & \approx 8.87 \end{align*}$$

(ii)

$$\begin{align} & \text{Class A performed better on average since the mean score of Class A (8) is higher than} \\ & \text{the mean score of Class B (7.875).} \\ \\ & \text{The scores obtained by students in Class A is more consistent since the standard deviation} \\ & \text{of Class A (2.60) is lower than the standard deviation of Class B (8.87).} \end{align}$$

(i)

$$\begin{align*} \text{Total score of 6 students} & = 10 \times 6 \\ & = 60 \\ \\ x + 5 + 16 + 6 + 10 + 4 & = 60 \\ x + 41 & = 60 \\ x & = 60 - 41 \\ x & = 19 \end{align*}$$

(ii)

$$\begin{align*} \sum f & = 6 \\ \\ \sum fx^2 & = 1 \times 5^2 + 1 \times 16^2 + 1 \times 6^2 + 1 \times 10^2 + 1 \times 4^2 + 1 \times 19^2 \\ & = 794 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {794 \over 6} - (10)^2 } \\ & = 5.6862 \\ & \approx 5.69 \end{align*}$$

(iii)

$$\begin{align} & \text{Priya is the best performer, since her score of 19 is much higher than the mean score of 10.} \end{align}$$

(i)

$$\begin{align*} \sum fx & = 1 \times 23 + 2 \times 15 + 1 \times 8 + 1 \times 13 + 1 \times 28 + 1 \times 6 \\ & = 108 \\ \\ \sum f & = 7 \\ \\ \sum fx^2 & = 1 \times 23^2 + 2 \times 15^2 + 1 \times 8^2 + 1 \times 13^2 + 1 \times 28^2 + 1 \times 6^2 \\ & = 2032 \\ \\ \text{Mean} & = {\sum fx \over \sum f} \\ & = {108 \over 7} \\ & = 15{3 \over 7} \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {2032 \over 7} - \left(15 {3 \over 7} \right)^2 } \\ & = 7.228 \\ & \approx 7.23 \end{align*}$$

(ii)

$$\begin{align*} \sum fx & = 1 \times 20 + 2 \times 12 + 1 \times 5 + 1 \times 10 + 1 \times 25 + 1 \times 3 \\ & = 87 \\ \\ \sum f & = 7 \\ \\ \sum fx^2 & = 1 \times 20^2 + 2 \times 12^2 + 1 \times 5^2 + 1 \times 10^2 + 1 \times 25^2 + 1 \times 3^2 \\ & = 1447 \\ \\ \text{Mean} & = {\sum fx \over \sum f} \\ & = {87 \over 7} \\ & = 12{3 \over 7} \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {1447 \over 7} - \left(12 {3 \over 7} \right)^2 } \\ & = 7.228 \\ & \approx 7.23 \end{align*}$$

(iii)

$$\begin{align} & \text{On average, Kate took a shorter time to fall asleep in New Zealand as the mean time taken is lesser.} \\ \\ & \text{The time taken to fall asleep in equally spread out in both locations as the standard deviation is the same.} \end{align}$$

(i)

$$\begin{align*} \text{For Train } A, \sum fx & = 3 \times 2 + 2 \times 3 + 5 \times 4 + 12 \times 5 + 10 \times 6 + 6 \times 7 + 1 \times 8 + 1 \times 9 \\ & = 211 \\ \\ \sum f & = 3 + 2 + 5 + 12 + 10 + 6 + 1 + 1 \\ & = 40 \\ \\ \sum fx^2 & = 3 \times 2^2 + 2 \times 3^2 + 5 \times 4^2 + 12 \times 5^2 + 10 \times 6^2 + 6 \times 7^2 + 1 \times 8^2 + 1 \times 9^2 \\ & = 1209 \\ \\ \text{Mean} & = {\sum fx \over \sum f} \\ & = {211 \over 40} \\ & = 5.275 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {1209 \over 40} - (5.275)^2 } \\ & = 1.5489 \\ & \approx 1.55 \\ \\ \\ \text{For Train } B, \sum fx & = 4 \times 2 + 3 \times 3 + 9 \times 4 + 9 \times 5 + 7 \times 6 + 5 \times 7 + 3 \times 8 + 0 \times 9 \\ & = 199 \\ \\ \sum f & = 4 + 3 + 9 + 9 + 7 + 5 + 3 + 0 \\ & = 40 \\ \\ \sum fx^2 & = 4 \times 2^2 + 3 \times 3^2 + 9 \times 4^2 + 9 \times 5^2 + 7 \times 6^2 + 5 \times 7^2 + 3 \times 8^2 + 0 \times 9^2 \\ & = 1101 \\ \\ \text{Mean} & = {\sum fx \over \sum f} \\ & = {199 \over 40} \\ & = 4.975 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {1101 \over 40} - (4.975)^2 } \\ & = 1.6656 \\ & \approx 1.67 \end{align*}$$

(ii)

$$\begin{align} & \text{Train A is more consistent since the standard deviation is lower.} \end{align}$$

(iii)

$$\begin{align} & \text{Train B is more punctual on the whole since the mean time by which the train arrived} \\ & \text{after the scheduled time was lower.} \end{align}$$

(a)(i) We need to use the mid-value for t: 21, 23, 25, 27, 29

$$\begin{align*} \text{Mean} & = {\sum fx \over \sum f} \\ & = {5 \times 21 + 11 \times 23 + 27 \times 25 + 13 \times 27 + 4 \times 29 \over 60} \\ & = 25 \end{align*}$$

(a)(ii)

$$\begin{align*} \sum fx^2 & = 5 \times 21^2 + 11 \times 23^2 + 27 \times 25^2 + 13 \times 27^2 + 4 \times 29^2 \\ & = 37740 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {37740 \over 60} - (25)^2 } \\ & = 2 \end{align*}$$

(b)

$$\begin{align} & \text{On the whole, the waiting time in both hospitals are similar since the mean waiting} \\ & \text{time is the same for both hospitals. } \\ \\ & \text{The waiting time is more consistent in Stamford Hospital as the standard deviation} \\ & \text{of the waiting time is lower for Stamford Hospital.} \end{align}$$

(a)(i) We need to use the mid-value for Temperature: 37.5, 42.5, 47.5, 52.5, 57.5, 62.5

$$\begin{align*} \text{Mean (City } A) & = {\sum fx \over \sum f} \\ & = {1 \times 37.5 + 4 \times 42.5 + 12 \times 47.5 + 23 \times 52.5 + 7 \times 57.5 + 3 \times 62.5 \over 50} \\ & = 51.5 \\ \\ \text{Mean (City } B) & = {2 \times 37.5 + 14 \times 42.5 + 16 \times 47.5 + 10 \times 52.5 + 5 \times 57.5 + 3 \times 62.5 \over 50} \\ & = 48.6 \end{align*}$$

(a)(ii)

$$\begin{align*} \text{For City } A, \sum fx^2 & = 1 \times 37.5^2 + 4 \times 42.5^2 + 12 \times 47.5^2 + 23 \times 52.5^2 + 7 \times 57.5^2 + 3 \times 62.5^2 \\ & = 133962.5 \\ \\ \text{S.D. (City } A) & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {133962.5 \over 50} - (51.5)^2 } \\ & = 5.19615 \\ & \approx 5.20 \\ \\ \\ \text{For City } B, \sum fx^2 & = 2 \times 37.5^2 + 14 \times 42.5^2 + 16 \times 47.5^2 + 10 \times 52.5^2 + 5 \times 57.5^2 + 3 \times 62.5^2 \\ & = 120012.5 \\ \\ \text{S.D. (City } B) & = \sqrt{ {120012.5 \over 50} - (48.6)^2 } \\ & = 6.18789 \\ & \approx 6.19 \end{align*}$$

(b)

$$\begin{align} & \text{City A as the mean temperature is higher.} \end{align}$$

(c)

$$\begin{align} & \text{City A as the standard deviation is lower.} \end{align}$$

$$\begin{align*} \text{Mean} & = {\sum fx \over \sum f} \\ & = {1 \times 10 + 1 \times 6 + 1 \times 18 + 1 \times x + 1 \times 15 + 1 \times y \over 6} \\ & = {10 + 6 + 18 + x + 15 + y \over 6} \\ & = {49 + x + y \over 6} \\ \\ 9 & = {49 + x + y \over 6} \\ 6(9) & = 49 + x + y \\ 54 & = 49 + x + y \\ 54 - 49 & = x + y \\ 5 & = x + y \\ \\ y & = 5 - x \phantom{000} \text{ --- (1)} \\ \\ \\ \sum fx^2 & = 1 \times 10^2 + 1 \times 6^2 + 1 \times 18^2 + 1 \times x^2 + 1 \times 15^2 + 1 \times y^2 \\ & = 100 + 36 + 324 + x^2 + 225 + y^2 \\ & = 685 + x^2 + y^2 \\ \\ \text{S.D.} & = \sqrt{ {\sum fx^2 \over \sum f} - (\text{Mean})^2 } \\ & = \sqrt{ {685 + x^2 + y^2 \over 6} - (9)^2 } \\ & = \sqrt{ {685 + x^2 + y^2 \over 6} - 81 } \\ \\ 6 & = \sqrt{ {685 + x^2 + y^2 \over 6} - 81} \\ 6^2 & = {685 + x^2 + y^2 \over 6} - 81 \\ 36 + 81 & = {685 + x^2 + y^2 \over 6} \\ 117 & = {685 + x^2 + y^2 \over 6} \\ 6(117) & = 685 + x^2 + y^2 \\ 702 & = 685 + x^2 + y^2 \\ 702 - 685 & = x^2 + y^2 \\ 17 & = x^2 + y^2 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 17 & = x^2 + (5 - x)^2 \\ 17 & = x^2 + 5^2 - 2(5)(x) + x^2 \\ 17 & = x^2 + 25 - 10x + x^2 \\ 0 & = 2x^2 - 10x + 25 - 17 \\ 0 & = 2x^2 - 10x + 8 \\ 0 & = x^2 - 5x + 4 \\ 0 & = (x - 1)(x - 4) \\ \\ x & = 1 \text{ or } x = 4 \\ \\ \text{Substitute } & \text{into (1),} \\ y & = 5 - 1 \text{ or } y = 5 - 4 \\ y & = 4 \phantom{000000} y = 1 \\ \\ \therefore x & = 1, y = 4 \text{ or } x = 4, y = 1 \end{align*}$$

(i)

$$\begin{align*} \text{Original mean} & = 5 \\ \\ \text{Total (for } n \text{ numbers)} & = 5 \times n \\ & = 5n \\ \\ \text{New mean} & = 5 \\ \\ \text{Total (for } n + 2 \text{ numbers)} & = 5 \times (n + 2) \\ & = 5(n + 2) \\ & = 5n + 10 \\ \\ \text{Sum of two additional numbers added} & = 5n + 10 - 5n \\ & = 10 \\ \\ \therefore & \phantom{0} \text{Sets } A \text{ and } C \end{align*}$$

(ii)

$$\begin{align} & \text{Set C} \\ \\ & \text{Since the mean is 5 and the original standard deviation is 1.8, majority of the data is around 3.2-6.8.} \\ \\ & \text{Since the values from Set C {4, 6} is within 3.2-6.8, the new standard deviation will be close to 1.8.} \end{align}$$

(i)

$$\begin{align*} \bar{x} & = {\sum fx \over \sum f} \\ & = {\sum fx \over 100} \\ \\ \bar{y} & = {\sum fy \over \sum f} \\ & = {\sum fy \over 100} \\ \\ {\bar{x} + \bar{y} \over 2} & = { {\sum fx \over 100} + {\sum fy \over 100} \over 2} \\ & = { {\sum fx + \sum fy \over 100} \over 2} \\ & = { \sum fx + \sum fy \over 200} \\ \\ \\ \text{Yes, } & {\bar{x} + \bar{y} \over 2} \text{ can be used} \end{align*}$$

(ii)

$$\begin{align} & \text{No. The combined standard deviation depends on the combined mean mass of both schools.} \end{align}$$

(iii)

$$\begin{align*} \sum fx & = 5 \times 45 + 36 \times 50 + 28 \times 55 + 22 \times 60 + 7 \times 65 + 2 \times 70 \\ & = 5480 \\ \\ \sum fy & = 7 \times 40 + 21 \times 45 + 24 \times 50 + 6 \times 55 + 3 \times 60 + 26 \times 65 + 8 \times 70 + 1 \times 75 + 4 \times 80 \\ & = 5580 \\ \\ \text{Mean} & = {5480 + 5580 \over 100 + 100} \\ & = 55.3 \\ \\ \\ \sum fx^2 & = 5 \times 45^2 + 36 \times 50^2 + 28 \times 55^2 + 22 \times 60^2 + 7 \times 65^2 + 2 \times 70^2 \\ & = 303400 \\ \\ \sum fy^2 & = 7 \times 40^2 + 21 \times 45^2 + 24 \times 50^2 + 6 \times 55^2 + 3 \times 60^2 + 26 \times 65^2 + 8 \times 70^2 + 1 \times 75^2 + 4 \times 80^2 \\ & = 322950 \\ \\ \text{S.D.} & = \sqrt{ {303400 + 322950 \over 100 + 100} - (55.3)^2 } \\ & = 8.5825 \\ & \approx 8.58 \end{align*}$$