Draw the cumulative frequency curve for the distribution.

here are the solutions to the problems.

10. Frequency Distribution

10. (a) Draw the cumulative frequency curve for the distribution.
To draw the cumulative frequency curve, we first need to calculate the cumulative frequencies and identify the upper class boundaries.

Step 1: Create a table with upper class boundaries and cumulative frequencies.

Marks (%)	Upper Class Boundary	Frequency	Cumulative Frequency
0-9	9.5	3	3
10-19	19.5	5	3 + 5 = 8
20-29	29.5	8	8 + 8 = 16
30-39	39.5	11	16 + 11 = 27
40-49	49.5	14	27 + 14 = 41
50-59	59.5	20	41 + 20 = 61
60-69	69.5	15	61 + 15 = 76
70-79	79.5	9	76 + 9 = 85
80-89	89.5	3	85 + 3 = 88
90-99	99.5	2	88 + 2 = 90

Step 2: Plot the cumulative frequency curve.

• Draw a graph with "Marks (%)" on the x-axis and "Cumulative Frequency" on the y-axis.
• Plot the points: $(9.5, 3)$, $(19.5, 8)$, $(29.5, 16)$, $(39.5, 27)$, $(49.5, 41)$, $(59.5, 61)$, $(69.5, 76)$, $(79.5, 85)$, $(89.5, 88)$, $(99.5, 90)$.
• Connect these points with a smooth curve. The curve should start from $(0,0)$ or the lower boundary of the first class with 0 cumulative frequency.

10. (b) Use your drawn graph to estimate:

10. (b) (i) the median mark
The median mark is the mark corresponding to the $50^{th}$ percentile of the total frequency.
Total frequency $N = 90$.
Median position = $\frac{N}{2} = \frac{90}{2} = 45^{th}$ value.

Step 1: On your cumulative frequency curve, locate 45 on the y-axis (cumulative frequency).

Step 2: Draw a horizontal line from 45 to intersect the curve.

Step 3: From the intersection point, draw a vertical line down to the x-axis (marks).
The value on the x-axis will be the median mark.
Based on the data, the median mark is approximately $\boxed{51.5\%}$. (This value is an estimate from interpolation, actual value depends on the drawn curve).

10. (b) (ii) the $60^{th}$ percentile mark
The $60^{th}$ percentile mark is the mark corresponding to $60\%$ of the total frequency.
Position of $60^{th}$ percentile = $\frac{60}{100} \times N = 0.60 \times 90 = 54^{th}$ value.

Step 1: On your cumulative frequency curve, locate 54 on the y-axis (cumulative frequency).

Step 2: Draw a horizontal line from 54 to intersect the curve.

Step 3: From the intersection point, draw a vertical line down to the x-axis (marks).
The value on the x-axis will be the $60^{th}$ percentile mark.
Based on the data, the $60^{th}$ percentile mark is approximately $\boxed{55.5\%}$. (This value is an estimate from interpolation, actual value depends on the drawn curve).

10. (c) If the pass mark was 50%, what percentage of students passed?
Students passed if their mark was 50% or higher. This corresponds to the classes 50-59, 60-69, 70-79, 80-89, and 90-99.

Step 1: Sum the frequencies for marks 50% and above.
Number of students who passed = Frequency (50-59) + Frequency (60-69) + Frequency (70-79) + Frequency (80-89) + Frequency (90-99)
Number of students who passed = $20 + 15 + 9 + 3 + 2 = 49$.

Step 2: Calculate the total number of students.
Total number of students $N = 90$.

Step 3: Calculate the percentage of students who passed.

$$\text{Percentage passed} = \frac{\text{Number of students who passed}}{\text{Total number of students}} \times 100\%$$ $$\text{Percentage passed} = \frac{49}{90} \times 100\%$$ $$\text{Percentage passed} \approx 54.44\%$$

The percentage of students who passed is approximately $\boxed{54.44\%}$.

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11. Probability

11. (a) A number is chosen at random from the set $F = \{3, 5, 6, 8, 11, 14, 17\}$.
The total number of elements in set $F$ is 7.

11. (a) (i) What is the probability that the number is even?

Step 1: Identify the even numbers in set $F$.
Even numbers: $\{6, 8, 14\}$.
Number of even numbers = 3.

Step 2: Calculate the probability.

$$P(\text{even}) = \frac{\text{Number of even numbers}}{\text{Total number of elements}} = \frac{3}{7}$$

The probability that the number is even is $\boxed{\frac{3}{7}}$.

11. (a) (ii) What is the probability that the number is prime?

Step 1: Identify the prime numbers in set $F$. (A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.)
Prime numbers: $\{3, 5, 11, 17\}$.
Number of prime numbers = 4.

Step 2: Calculate the probability.

$$P(\text{prime}) = \frac{\text{Number of prime numbers}}{\text{Total number of elements}} = \frac{4}{7}$$

The probability that the number is prime is $\boxed{\frac{4}{7}}$.

11. (a) (iii) What is the probability that the number is greater than 10?

Step 1: Identify the numbers greater than 10 in set $F$.
Numbers greater than 10: $\{11, 14, 17\}$.
Number of elements greater than 10 = 3.

Step 2: Calculate the probability.

$$P(\text{greater than 10}) = \frac{\text{Number of elements greater than 10}}{\text{Total number of elements}} = \frac{3}{7}$$

The probability that the number is greater than 10 is $\boxed{\frac{3}{7}}$.

11. (a) (iv) What is the probability that the number is a multiple of 3?

Step 1: Identify the multiples of 3 in set $F$.
Multiples of 3: $\{3, 6\}$.
Number of multiples of 3 = 2.

Step 2: Calculate the probability.

$$P(\text{multiple of 3}) = \frac{\text{Number of multiples of 3}}{\text{Total number of elements}} = \frac{2}{7}$$

The probability that the number is a multiple of 3 is $\boxed{\frac{2}{7}}$.

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11. (b) A bag contains 6 blue balls, 7 green balls and 3 yellow balls of the same size.
Total number of balls = $6 + 7 + 3 = 16$.
One ball is drawn at random from the bag and not replaced. A second ball is then drawn.

11. (b) (i) Find the probability of obtaining two yellow balls.

Step 1: Probability of drawing the first yellow ball.
There are 3 yellow balls out of 16 total.

$$P(\text{1st yellow}) = \frac{3}{16}$$

Step 2: Probability of drawing the second yellow ball (given the first was yellow and not replaced).
Now there are 2 yellow balls left and 15 total balls.

$$P(\text{2nd yellow | 1st yellow}) = \frac{2}{15}$$

Step 3: Calculate the probability of drawing two yellow balls.

$$P(\text{two yellow balls}) = P(\text{1st yellow}) \times P(\text{2nd yellow | 1st yellow})$$ $$P(\text{two yellow balls}) = \frac{3}{16} \times \frac{2}{15} = \frac{6}{240} = \frac{1}{40}$$

The probability of obtaining two yellow balls is $\boxed{\frac{1}{40}}$.

11. (b) (ii) Find the probability of obtaining two blue balls or two green balls.

Step 1: Probability of drawing two blue balls (BB).

$$P(\text{1st blue}) = \frac{6}{16}$$ $$P(\text{2nd blue | 1st blue}) = \frac{5}{15}$$ P(BB) = \frac{6}{16} \times \