Past Exam Questions

To find \(f\) and the total number of children:

The midpoints of the intervals are: 5, 15, 30, and 60.

The equation for the mean is:

\(\frac{5 \times 2 + 15f + 30 \times 11 + 60 \times 4}{17 + f} = 27.5\)

\(10 + 15f + 330 + 240 = 27.5(17 + f)\)

\(15f + 580 = 467.5 + 27.5f\)

\(580 - 467.5 = 27.5f - 15f\)

\(112.5 = 12.5f\)

\(f = 9\)

Total number of children = \(17 + f = 26\).

To find the standard deviation:

Calculate \(\sigma\) using the formula:

\(\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{n}}\)

\(\sigma = 16.1\)

(i) To find the mean:

\(\bar{x} = \frac{\Sigma x}{n} = \frac{745}{18} = 41.4\)

To find the standard deviation:

\(s = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} = \sqrt{\frac{33951}{18} - \left(\frac{745}{18}\right)^2} = 13.2\)

(ii) Let the age of the person who left be \(x\).

The sum of the ages of the remaining 17 people is:

\(17 \times 41 = 697\)

Therefore, \(745 - x = 697\), giving \(x = 48\).

To find the standard deviation of the remaining 17 people:

\(s = \sqrt{\frac{\Sigma x^2 - x^2}{17} - (41)^2} = \sqrt{\frac{33951 - 48^2}{17} - 41^2} = 13.4\)

(i) To find the mean for Team A:

Mean, \(\bar{x}_A = \frac{150 + 220 + 77 + 30 + 298 + 118 + 160 + 57}{8} = 139\).

To find the standard deviation for Team A:

\(First, calculate the variance:\)

Variance, \(s_A^2 = \frac{(150-139)^2 + (220-139)^2 + (77-139)^2 + (30-139)^2 + (298-139)^2 + (118-139)^2 + (160-139)^2 + (57-139)^2}{8}\).

\(s_A^2 = \frac{121 + 6561 + 3844 + 11881 + 25281 + 441 + 441 + 6724}{8} = 6901.25\).

Standard deviation, \(\sigma_A = \sqrt{6901.25} = 83.1\).

(ii) Team B has a smaller standard deviation (29.63) compared to Team A (83.1), indicating that Team B's scores are more consistent.

Let the random variable \(X\) take values 0 and 2 with frequencies 23 and 17 respectively.

The total number of observations is 40.

The mean \(\bar{x}\) is calculated as:

\(\bar{x} = \frac{(0 \times 23) + (2 \times 17)}{40} = \frac{34}{40} = 0.85\)

The variance \(s^2\) is calculated using:

\(s^2 = \frac{(0^2 \times 23) + (2^2 \times 17)}{40} - (\bar{x})^2\)

\(s^2 = \frac{(0 \times 23) + (4 \times 17)}{40} - (0.85)^2\)

\(s^2 = \frac{68}{40} - 0.7225\)

\(s^2 = 1.7 - 0.7225 = 0.9775\)

\(Thus, the variance is approximately 0.978.\)

(a) To find the mean, sum all the estimates and divide by the number of estimates:

\(\text{Mean} = \frac{4.1 + 4.2 + 4.4 + 4.5 + 4.6 + 4.8 + 5.0 + 5.2 + 5.3 + 5.4 + 5.5 + 5.8 + 6.0 + 6.2 + 6.3 + 6.4 + 6.6 + 6.8 + 6.9 + 19.4}{20} = \frac{123.4}{20} = 6.17\)

(b) To find the median, arrange the data in ascending order and find the middle value. Since there are 20 values, the median is the average of the 10th and 11th values:

\(\text{Median} = \frac{5.4 + 5.5}{2} = 5.45\)

(c) The median is more suitable because the mean is unduly influenced by the extreme value of 19.4, which skews the average higher than the central tendency of the majority of the data.

(a) To find the mean height of all 17 members, we use the formula for the mean:

\(\text{Mean height} = \frac{\Sigma x + \Sigma y}{6 + 11} = \frac{1050 + 1991}{17} = \frac{3041}{17} = 178.9 \text{ cm}\)

\((b) To find the standard deviation, we first calculate the variance:\)

\(\Sigma x^2 + \Sigma y^2 = 193700 + 366400 = 560100\)

\(\text{Variance} = \frac{560100}{17} - \left(\frac{3041}{17}\right)^2 = \frac{560100}{17} - 178.88^2\)

\(\text{Variance} = 32947.0588 - 32008.289 = 938.7698\)

\(\text{Standard deviation} = \sqrt{938.7698} = 30.8 \text{ cm}\)

To find the median and interquartile range, first arrange the data in ascending order: 30, 38, 40, 40, 45, 50, 52, 55, 55, 60, 62, 110.

Median: The median is the average of the 6th and 7th values: \(\frac{50 + 52}{2} = 51\).

Interquartile Range (IQR): The lower quartile \(Q_1\) is the 3rd value: 40. The upper quartile \(Q_3\) is the 9th value: 57.5. Thus, \(\text{IQR} = Q_3 - Q_1 = 57.5 - 40 = 17.5\).

Disadvantage of using the mean: The mean is disproportionately affected by the extreme value 110, which is an outlier.

Given: Mean of 20 values \(\bar{x} = 60\) and standard deviation \(s = 4\).

(i) To find \(\Sigma x\) and \(\Sigma x^2\):

\(\Sigma x = 60 \times 20 = 1200\)

\(Using the formula for variance:\)

\(s^2 = \frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2\)

\(4^2 = \frac{\Sigma x^2}{20} - 60^2\)

\(16 = \frac{\Sigma x^2}{20} - 3600\)

\(\frac{\Sigma x^2}{20} = 3616\)

\(\Sigma x^2 = 3616 \times 20 = 72320\)

(ii) For the combined 30 values:

\(\Sigma x = 1200 + 550 = 1750\)

\(\Sigma x^2 = 72320 + 40500 = 112800\)

Mean of 30 values:

\(\bar{x} = \frac{1750}{30} = 58.3\)

Variance of 30 values:

\(s^2 = \frac{112800}{30} - \left(\frac{1750}{30}\right)^2\)

\(s^2 = 3760 - 3412.89 = 357.11\)

Standard deviation:

\(s = \sqrt{357.11} \approx 18.9\)

To determine the most suitable measure of central tendency, we consider the mean, mode, and median:

• Mean: The mean is affected by extreme values. In this dataset, the value 768 is much larger than the others, skewing the mean upwards. Therefore, the mean is not suitable.
• Mode: The mode is the most frequently occurring value. In this dataset, all values are different, so there is no mode. Thus, the mode is not suitable.
• Median: The median is the middle value when the data is ordered. It is not affected by extreme values. Ordering the data: 180, 194, 227, 228, 229, 235, 238, 242, 246, 275, 311, 332, 768. The median is 238, which is a representative measure of central tendency for this dataset.

Therefore, the median is the most suitable measure of central tendency for these times.

First, calculate the new sums:

\(\Sigma x = 1923 + 166 + 172 + 182 = 2443\)

\(\Sigma x^2 = 337221 + 166^2 + 172^2 + 182^2 = 427485\)

Calculate the mean:

\(\text{Mean} = \frac{\Sigma x}{14} = \frac{2443}{14} = 174.5\)

Calculate the variance:

\(\text{Variance} = \frac{\Sigma x^2}{14} - \left(\frac{\Sigma x}{14}\right)^2 = \frac{427485}{14} - \left(\frac{2443}{14}\right)^2\)

\(= 30534.64 - 30480.25 = 54.39\)

Calculate the standard deviation:

\(\text{Standard deviation} = \sqrt{54.39} = 9.19\)

(a) The graph of \(y = |4x - 2|\) is a V-shaped graph. The vertex of the graph is at \(x = \frac{1}{2}\), \(y = 0\). The graph is symmetrical about the vertical line \(x = \frac{1}{2}\) and extends into the second quadrant.

(b) To solve \(1 + 3x < |4x - 2|\), consider two cases:

Case 1: \(4x - 2 \geq 0\) (i.e., \(x \geq \frac{1}{2}\))

Then \(|4x - 2| = 4x - 2\).

Solving \(1 + 3x < 4x - 2\) gives:

\(1 + 3x < 4x - 2\)

\(1 + 3x - 4x < -2\)

\(1 - x < -2\)

\(-x < -3\)

\(x > 3\)

Case 2: \(4x - 2 < 0\) (i.e., \(x < \frac{1}{2}\))

Then \(|4x - 2| = -(4x - 2) = -4x + 2\).

Solving \(1 + 3x < -4x + 2\) gives:

\(1 + 3x < -4x + 2\)

\(1 + 3x + 4x < 2\)

\(1 + 7x < 2\)

\(7x < 1\)

\(x < \frac{1}{2}\)

Thus, the solution is \(x < \frac{1}{2}\) or \(x > 3\).

To solve the inequality \(|3x - a| > 2|x + 2a|\), we first consider the non-modular inequality:

\((3x - a)^2 > 2^2(x + 2a)^2\).

This simplifies to:

\((3x - a)^2 > 4(x + 2a)^2\).

Expanding both sides, we get:

\(9x^2 - 6ax + a^2 > 4(x^2 + 4ax + 4a^2)\).

Simplifying further:

\(9x^2 - 6ax + a^2 > 4x^2 + 16ax + 16a^2\).

Rearranging terms gives:

\(5x^2 - 22ax - 15a^2 > 0\).

We solve the quadratic equation \(5x^2 - 22ax - 15a^2 = 0\) to find critical values:

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -22a\), \(c = -15a^2\), we find:

\(x = \frac{22a \pm \sqrt{(-22a)^2 - 4 \cdot 5 \cdot (-15a^2)}}{2 \cdot 5}\).

\(x = \frac{22a \pm \sqrt{484a^2 + 300a^2}}{10}\).

\(x = \frac{22a \pm \sqrt{784a^2}}{10}\).

\(x = \frac{22a \pm 28a}{10}\).

This gives critical values \(x = 5a\) and \(x = -\frac{3}{5}a\).

The solution to the inequality is:

\(x > 5a\) or \(x < -\frac{3}{5}a\).

To solve the inequality \(|2x - 1| < 3|x + 1|\), we first consider the non-modular inequality \((2x - 1)^2 < 3^2(x + 1)^2\).

Expanding both sides, we get:

\((2x - 1)^2 = 4x^2 - 4x + 1\)

\(3^2(x + 1)^2 = 9(x^2 + 2x + 1) = 9x^2 + 18x + 9\)

Setting up the inequality:

\(4x^2 - 4x + 1 < 9x^2 + 18x + 9\)

Rearranging terms gives:

\(4x^2 - 4x + 1 - 9x^2 - 18x - 9 < 0\)

\(-5x^2 - 22x - 8 < 0\)

Solving the quadratic equation \(-5x^2 - 22x - 8 = 0\) gives the critical values \(x = -4\) and \(x = -\frac{2}{5}\).

The solution to the inequality is \(x < -4\) or \(x > -\frac{2}{5}\).

To solve the inequality \(2|3x - 1| < |x + 1|\), we first consider the critical points where the expressions inside the absolute values are zero: \(3x - 1 = 0\) and \(x + 1 = 0\).

Solving these gives \(x = \frac{1}{3}\) and \(x = -1\). These points divide the number line into intervals.

Consider the intervals: \((-\infty, -1)\), \((-1, \frac{1}{3})\), and \((\frac{1}{3}, \infty)\).

For each interval, remove the absolute values by considering the sign of the expressions:

1. For \(x < -1\), both \(3x - 1\) and \(x + 1\) are negative, so the inequality becomes \(2(-(3x - 1)) < -(x + 1)\).

2. For \(-1 < x < \frac{1}{3}\), \(3x - 1\) is negative and \(x + 1\) is positive, so the inequality becomes \(2(-(3x - 1)) < x + 1\).

3. For \(x > \frac{1}{3}\), both \(3x - 1\) and \(x + 1\) are positive, so the inequality becomes \(2(3x - 1) < x + 1\).

Solving these inequalities gives the critical values \(x = \frac{3}{5}\) and \(x = \frac{1}{7}\).

The solution is \(\frac{1}{7} < x < \frac{3}{5}\).

To solve the inequality \(2 - 5x > 2|x - 3|\), we consider the expression inside the absolute value, \(|x - 3|\), which can be split into two cases:

Case 1: \(x - 3 \geq 0\) (i.e., \(x \geq 3\))

In this case, \(|x - 3| = x - 3\). The inequality becomes:

\(2 - 5x > 2(x - 3)\)

\(2 - 5x > 2x - 6\)

\(2 + 6 > 2x + 5x\)

\(8 > 7x\)

\(x < \frac{8}{7}\)

However, since \(x \geq 3\), there is no solution in this case.

Case 2: \(x - 3 < 0\) (i.e., \(x < 3\))

In this case, \(|x - 3| = -(x - 3) = -x + 3\). The inequality becomes:

\(2 - 5x > 2(-x + 3)\)

\(2 - 5x > -2x + 6\)

\(2 - 6 > -2x + 5x\)

\(-4 > 3x\)

\(x < -\frac{4}{3}\)

Thus, the solution is \(x < -\frac{4}{3}\).

To solve the inequality \(|2x - 1| > 3|x + 2|\), we first consider the non-modular inequality:

\((2x - 1)^2 > 3^2(x + 2)^2\)

Expanding both sides, we have:

\((2x - 1)^2 = 4x^2 - 4x + 1\)

\(9(x + 2)^2 = 9(x^2 + 4x + 4) = 9x^2 + 36x + 36\)

Setting up the inequality:

\(4x^2 - 4x + 1 > 9x^2 + 36x + 36\)

Rearranging terms gives:

\(4x^2 - 4x + 1 - 9x^2 - 36x - 36 > 0\)

\(-5x^2 - 40x - 35 > 0\)

Dividing the entire inequality by -1 (and reversing the inequality sign):

\(5x^2 + 40x + 35 < 0\)

Factoring the quadratic:

\((x + 7)(x + 1) < 0\)

The critical points are \(x = -7\) and \(x = -1\).

Testing intervals, we find the solution is:

-7 < x < -1

To solve the inequality \(|x - 2| < 3x - 4\), we first consider the expression inside the absolute value and the linear expression on the right.

1. Find the x-coordinate of intersection by solving the equation \(x - 2 = 3x - 4\).

2. Rearrange to get \(x - 3x = -4 + 2\), which simplifies to \(-2x = -2\).

3. Solving for \(x\), we get \(x = 1\).

4. Consider the inequality \(3x - 4 > 2 - x\) to find the critical value.

5. Rearrange to get \(3x + x > 4 + 2\), which simplifies to \(4x > 6\).

6. Solving for \(x\), we get \(x > \frac{3}{2}\).

Thus, the solution to the inequality is \(x > \frac{3}{2}\).

The function \(y = |x - 2|\) is an absolute value function. The graph of \(y = |x - 2|\) is a V-shaped graph.

The vertex of the graph is at the point \((2, 0)\) because the expression inside the absolute value, \(x - 2\), equals zero when \(x = 2\).

For \(x < 2\), the graph is a line with a negative slope, \(y = -(x - 2) = -x + 2\).

For \(x > 2\), the graph is a line with a positive slope, \(y = x - 2\).

The graph is symmetric about the vertical line \(x = 2\).

To solve the inequality \(|2x - 3| > 4|x + 1|\), we first consider the non-modular inequality:

\((2x - 3)^2 > 4^2(x + 1)^2\)

This simplifies to:

\((2x - 3)^2 > 16(x + 1)^2\)

Expanding both sides, we get:

\(4x^2 - 12x + 9 > 16(x^2 + 2x + 1)\)

\(4x^2 - 12x + 9 > 16x^2 + 32x + 16\)

Rearranging terms gives:

\(4x^2 - 12x + 9 - 16x^2 - 32x - 16 < 0\)

\(-12x^2 - 44x - 7 < 0\)

Solving the quadratic inequality \(12x^2 + 44x + 7 < 0\) gives the critical values:

\(x = -\frac{7}{2}\) and \(x = -\frac{1}{6}\)

The solution to the inequality is:

\(\frac{7}{2} < x < -\frac{1}{6}\)

To solve the inequality \(3|2x - 1| > |x + 4|\), we first consider the non-modular inequality \(3^2(2x - 1)^2 > (x + 4)^2\).

This simplifies to \(35x^2 - 44x - 7 = 0\).

We solve the quadratic equation \(35x^2 - 44x - 7 = 0\) by factoring or using the quadratic formula.

Factoring gives \((5x - 7)(7x + 1) = 0\), leading to critical values \(x = \frac{7}{5}\) and \(x = -\frac{1}{7}\).

The solution to the inequality is \(x > \frac{7}{5}\) or \(x < -\frac{1}{7}\).