Mastery Guide: Averages, Bar & Line Graphs, Pie Charts & Theoretical Probability

(a) $12\text{ kg}, 16\text{ kg}, 26\text{ kg}, 42\text{ kg}$:
$$\text{Sum} = 12 + 16 + 26 + 42 = 96\text{ kg}, \quad \text{Count } n = 4$$ $$\text{Average} = \frac{96}{4} = \mathbf{24\text{ kg}}$$

(b) $10\text{ cm}, 13\text{ cm}, 17\text{ cm}, 16\text{ cm}, 19\text{ cm}$:
$$\text{Sum} = 10 + 13 + 17 + 16 + 19 = 75\text{ cm}, \quad \text{Count } n = 5$$ $$\text{Average} = \frac{75}{5} = \mathbf{15\text{ cm}}$$

(c) $5\text{ l}, 15\text{ l}, 30\text{ l}, 25\text{ l}, 40\text{ l}$:
$$\text{Sum} = 5 + 15 + 30 + 25 + 40 = 115\text{ l}, \quad \text{Count } n = 5$$ $$\text{Average} = \frac{115}{5} = \mathbf{23\text{ l}}$$

(d) $60\text{ m}, 48\text{ m}, 52\text{ m}, 64\text{ m}$:
$$\text{Sum} = 60 + 48 + 52 + 64 = 224\text{ m}, \quad \text{Count } n = 4$$ $$\text{Average} = \frac{224}{4} = \mathbf{56\text{ m}}$$

Steps to Construct:

• Draw the vertical axis and list the fruits: Strawberry, Apple, Banana, Mango.
• Draw the horizontal axis with scale $0$ to $10$ with unit step of $1$.
• Draw horizontal bars: Strawberry to $5$, Apple to $4$, Banana to $8$, Mango to $9$.

Steps to Construct:

• Horizontal axis: Months (Feb, Mar, Apr, May, Jun).
• Vertical axis: Number of patients (Scale: $0, 2, 4, 6, 8, 10, 12, 14, 16$).
• Draw vertical bars: Feb ($2$), Mar ($6$), Apr ($10$), May ($12$), Jun ($15$).

Steps to Construct:

• Horizontal axis: Animal names.
• Vertical axis: Number of animals (Scale: $0, 2, 4, 6, 8, 10, 12, 14$).
• Draw bars corresponding to each animal's count.

Steps to Construct:

• Vertical axis: Days of the week (Mon to Sun).
• Horizontal axis: Temperature in $^\circ\text{C}$ (Scale: $0, 5, 10, 15, 20, 25, 30, 35, 40$).
• Draw horizontal bars extending to the given degree on each day.

(a) Which mode of transport is used by the most students?
→ Bus ($100\text{ students}$)

(b) Which mode of transport is used by the fewest students?
→ Motorbike ($20\text{ students}$)

(c) Find the difference between the highest and lowest number of students:
→ $100 - 20 = \mathbf{80\text{ students}}$

(d) How many students use rickshaws to come to school?
→ $60\text{ students}$

(e) Find the total number of students in the survey:
→ $100 + 60 + 70 + 20 + 80 = \mathbf{330\text{ students}}$

(a) Which pet animal is kept by the most students?
→ Cat ($9\text{ students}$)

(b) Which pet animal is kept by the fewest students?
→ Cow ($0\text{ students}$)

(c) Find the total number of pets kept by students:
→ $9 + 4 + 7 + 0 = \mathbf{20\text{ pets}}$

(d) How many students do not keep any pet animal?
→ $\text{Total students} - \text{Students with pets} = 35 - 20 = \mathbf{15\text{ students}}$

(e) How many more students keep buffaloes than goats?
→ $7 - 4 = \mathbf{3\text{ more students}}$

(a) On which day was the number of customers maximum?
→ Sunday ($1\text{ }000\text{ customers}$)

(b) On which day was the number of customers minimum?
→ Tuesday ($300\text{ customers}$)

(c) Is the sum of customers on Monday and Thursday greater than Sunday?
→ $\text{Mon} + \text{Thu} = 700 + 400 = 1\text{ }100$. Yes, $1\text{ }100 > 1\text{ }000$, so it is Greater.

(d) How many customers visited the store on Friday?
→ $600\text{ customers}$

(a) On which day was the temperature highest?
→ Monday ($38^\circ\text{C}$)

(b) Which two days had the same temperature?
→ Thursday and Friday ($35^\circ\text{C}$ each)

(c) On which day was the temperature lowest?
→ Wednesday ($30^\circ\text{C}$)

(d) Find the temperature difference between Monday and Wednesday:
→ $38^\circ\text{C} - 30^\circ\text{C} = \mathbf{8^\circ\text{C}}$

(a) How much Cholesterol is in the boiled egg?
→ $187\text{ mg}$

(b) Which is the largest nutritional component by weight (in grams)?
→ Proteins ($6\text{ g}$)

(c) How much Potassium is present?
→ $63\text{ mg}$

(d) Find the total mass of proteins and fat together:
→ $6\text{ g} + 5\text{ g} = \mathbf{11\text{ g}}$

(e) Find the difference between Potassium and Sodium:
→ $63\text{ mg} - 62\text{ mg} = \mathbf{1\text{ mg}}$

(a) Find the total number of cattle in the village:
→ $98 + 35 + 74 + 100 + 58 = \mathbf{365\text{ cattle}}$

(b) Which animal is most bred, and which is least bred?
→ Most bred: Goats ($100$) • Least bred: Donkeys ($35$)

(c) Compare the sum of sheep and goats with the sum of donkeys and buffaloes:
→ $\text{Sheep} + \text{Goats} = 58 + 100 = 158$
→ $\text{Donkeys} + \text{Buffaloes} = 35 + 98 = 133$
→ Difference $= 158 - 133 = \mathbf{25\text{ more sheep and goats}}$

(d) What fraction of total cattle are Bulls?
→ $$\text{Fraction} = \mathbf{\frac{74}{365}}$$

Sample space: $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \implies n(S) = 10$.

(i) Getting 4: Favorable: $\{4\} \implies P(4) = \mathbf{\frac{1}{10}}$

(ii) Getting a number less than 4: Favorable: $\{1, 2, 3\} \implies P(<4) = \mathbf{\frac{3}{10}}$

(iii) Getting a number greater than 4: Favorable: $\{5, 6, 7, 8, 9, 10\} \implies P(>4) = \frac{6}{10} = \mathbf{\frac{3}{5}}$

(i) Bag A has 8 red and 4 green balls. Find the probability of drawing a green ball:
$$n(S) = 8 + 4 = 12, \quad n(\text{green}) = 4 \implies P(\text{green}) = \frac{4}{12} = \mathbf{\frac{1}{3}}$$

(ii) Bag B has 7 black and 5 orange balls. Find the probability of drawing a black ball:
$$n(S) = 7 + 5 = 12, \quad n(\text{black}) = 7 \implies P(\text{black}) = \mathbf{\frac{7}{12}}$$

Total cards: $n(S) = 60 - 11 + 1 = 50$.

(i) A multiple of 10: Multiples are $\{20, 30, 40, 50, 60\} \implies n(E) = 5$.
$$P = \frac{5}{50} = \mathbf{\frac{1}{10}}$$

(ii) A multiple of 5: Multiples are $\{15, 20, 25, 30, 35, 40, 45, 50, 55, 60\} \implies n(E) = 10$.
$$P = \frac{10}{50} = \mathbf{\frac{1}{5}}$$

Sample space: $S = \{1, 2, ..., 20\} \implies n(S) = 20$.

(i) A multiple of 3: $\{3, 6, 9, 12, 15, 18\} \implies n = 6 \implies P = \frac{6}{20} = \mathbf{\frac{3}{10}}$

(ii) A multiple of 5: $\{5, 10, 15, 20\} \implies n = 4 \implies P = \frac{4}{20} = \mathbf{\frac{1}{5}}$

(iii) An odd number: $\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\} \implies n = 10 \implies P = \frac{10}{20} = \mathbf{\frac{1}{2}}$

(iv) A composite number: $\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20\} \implies n = 11 \implies P = \mathbf{\frac{11}{20}}$

(v) A prime number: $\{2, 3, 5, 7, 11, 13, 17, 19\} \implies n = 8 \implies P = \frac{8}{20} = \mathbf{\frac{2}{5}}$

(vi) Neither prime nor composite ($1$): $\{1\} \implies n = 1 \implies P = \mathbf{\frac{1}{20}}$

Sample space: $S = \{1, 2, 3, 4, 5, 6, 7, 8\} \implies n(S) = 8$.

(i) An odd number: $\{1, 3, 5, 7\} \implies n = 4 \implies P = \frac{4}{8} = \mathbf{\frac{1}{2}}$

(ii) An even number: $\{2, 4, 6, 8\} \implies n = 4 \implies P = \frac{4}{8} = \mathbf{\frac{1}{2}}$

(iii) A multiple of 4: $\{4, 8\} \implies n = 2 \implies P = \frac{2}{8} = \mathbf{\frac{1}{4}}$

(iv) Number 10: Impossible event $\implies P(10) = \frac{0}{8} = \mathbf{0}$

(v) Number 0: Impossible event $\implies P(0) = \frac{0}{8} = \mathbf{0}$

Total cards: $n(S) = 10 + 12 + 15 + 9 = 46$.

(i) A red card: $P(\text{red}) = \frac{10}{46} = \mathbf{\frac{5}{23}}$

(ii) A green card: $P(\text{green}) = \mathbf{\frac{15}{46}}$

(iii) A red or orange card: Favorable: $10 + 9 = 19 \implies P = \mathbf{\frac{19}{46}}$

(a) Average is equal to: → $\frac{\text{Sum of quantities}}{\text{Number of quantities}}$

(b) Average of $19, 21, 22, 24, 19$ is: → $\frac{105}{5} = \mathbf{21}$

(c) Total sum of quantities is equal to: → $\text{Average} \times \text{Number of quantities}$

(d) If total value is 600 and average is 50, then number of quantities is: → $\frac{600}{50} = \mathbf{12}$

(e) Bisma made 3 long jumps of $12\text{ cm}, 9\text{ cm}, 15\text{ cm}$. Her average jump is: → $\frac{12 + 9 + 15}{3} = \frac{36}{3} = \mathbf{12\text{ cm}}$

First 6 Odd Numbers: $1, 3, 5, 7, 9, 11 \implies \text{Sum} = 36 \implies \text{Average} = \frac{36}{6} = \mathbf{6}$

First 6 Even Numbers: $2, 4, 6, 8, 10, 12 \implies \text{Sum} = 42 \implies \text{Average} = \frac{42}{6} = \mathbf{7}$

Comparison: The average of even numbers ($7$) is greater than the average of odd numbers ($6$) by $7 - 6 = \mathbf{1}$.

Vertical Bar Graph: Months on $x$-axis; Profit on $y$-axis (Scale: $0, 10000, 20000, ..., 80000$).

(i) Probability of getting a Head: $P(\text{Head}) = \frac{75}{125} = \frac{3}{5} = \mathbf{0.6}$

(ii) Probability of getting a Tail: $P(\text{Tail}) = \frac{50}{125} = \frac{2}{5} = \mathbf{0.4}$

(i) Probability of getting a defective bulb: $P(\text{defective}) = \frac{10}{25} = \frac{2}{5} = \mathbf{0.4}$

(ii) Probability of getting a non-defective bulb: Non-defective $= 25 - 10 = 15 \implies P = \frac{15}{25} = \frac{3}{5} = \mathbf{0.6}$

(a) Which is the most favourite subject? → Arts ($10\text{ students}$)

(b) Which is the least favourite subject? → Social Studies ($3\text{ students}$)

(c) Find total number of students: → $5 + 6 + 10 + 6 + 8 + 3 = \mathbf{38\text{ students}}$

(d) How many students like Science and Mathematics altogether? → $6 + 8 = \mathbf{14\text{ students}}$

(e) Which two subjects are liked by the same number of students? → English and Science ($6\text{ students each}$)

(f) How many more students like Mathematics than Science? → $8 - 6 = \mathbf{2\text{ students}}$