Past Exam Questions

To solve the problem, we first organize the data in ascending order: 36, 37, 38, 38, 38, 39, 40, 40, 42, 45.

1. Mode: The mode is the number that appears most frequently. Here, 38 appears three times, which is more frequent than any other number. Thus, the mode is 38.
2. Median: The median is the middle value of the ordered data. Since there are 10 data points, the median is the average of the 5th and 6th values: \(\frac{38 + 39}{2} = 38.5\).
3. Interquartile Range (IQR): The IQR is the difference between the upper quartile \(Q_3\) and the lower quartile \(Q_1\).
   • \(Q_1\) is the median of the first half of the data: 36, 37, 38, 38, 38. The median is 38.
   • \(Q_3\) is the median of the second half of the data: 39, 40, 40, 42, 45. The median is 40.
   • Thus, \(\text{IQR} = Q_3 - Q_1 = 40 - 38 = 2\).

(i) To find the mean age of all 19 people, use the formula for the combined mean:

\(\bar{x} = \frac{(48.7 \times 12) + (38.1 \times 7)}{19} = \frac{584.4 + 266.7}{19} = \frac{851.1}{19} = 44.8 \text{ years}\)

(ii) First, find \(\Sigma x^2\) and \(\Sigma y^2\) using the given standard deviations and means:

For the first group:

\(7.65^2 = \frac{\Sigma x^2}{12} - 48.7^2\)

\(58.5225 = \frac{\Sigma x^2}{12} - 2371.69\)

\(\Sigma x^2 = 12 \times (58.5225 + 2371.69) = 29162.55\)

For the second group:

\(4.2^2 = \frac{\Sigma y^2}{7} - 38.1^2\)

\(17.64 = \frac{\Sigma y^2}{7} - 1451.61\)

\(\Sigma y^2 = 7 \times (17.64 + 1451.61) = 10284.75\)

\(Now, find the combined variance:\)

\(\text{Combined variance} = \frac{29162.55 + 10284.75}{19} - 44.79^2\)

\(= \frac{39447.3}{19} - 2006.0641\)

\(= 2076.1737 - 2006.0641 = 70.1096\)

Finally, the standard deviation is:

\(\sigma = \sqrt{70.1096} \approx 8.37 \text{ or } 8.36 \text{ years}\)

(i) Let the cost of a sweater be \(x\).

The total cost of the items is given by:

\(4 \times 5.5 + 3x + 90 = 8 \times 29\)

\(22 + 3x + 90 = 232\)

\(112 + 3x = 232\)

\(3x = 120\)

\(x = 40\)

Thus, the cost of a sweater is $40.

(ii) Since the standard deviation is $0, all items cost the same. Therefore, the cost of each item is $26. Diksha buys 4 shirts, so the total cost for the shirts is:

\(4 \times 26 = 104\)

Thus, Diksha spends $104 on shirts.

To estimate the mean, we first find the midpoints of each class interval:

• 1 – 25: midpoint = 13
• 26 – 35: midpoint = 30.5
• 36 – 45: midpoint = 40.5
• 46 – 55: midpoint = 50.5
• 56 – 90: midpoint = 73

Next, we calculate the mean using the formula:

\(\bar{x} = \frac{\sum f x}{\sum f} = \frac{4 \times 13 + 24 \times 30.5 + 38 \times 40.5 + 34 \times 50.5 + 20 \times 73}{120} = \frac{5500}{120} = 45.8\)

\(To find the variance, we use:\)

\(s^2 = \frac{\sum f x^2}{\sum f} - \bar{x}^2\)

\(\sum f x^2 = 4 \times 13^2 + 24 \times 30.5^2 + 38 \times 40.5^2 + 34 \times 50.5^2 + 20 \times 73^2 = 278620\)

\(s^2 = \frac{278620}{120} - 45.8^2 = 2321.8333 - 2097.64 = 224.1933\)

The standard deviation is:

\(s = \sqrt{224.1933} = 14.9\)

(i) To find the mean weight of the 27 birds, use the formula for the combined mean:

\(\text{new mean} = \frac{9 \times 7.1 + 18 \times 5.2}{27} = 5.83 \text{ kg}\)

(ii) First, find \(\Sigma x_t^2\) and \(\Sigma x_g^2\) using the variance formula:

For turkeys: \(1.45^2 = \frac{\Sigma x_t^2}{9} \Rightarrow \Sigma x_t^2 = 472.6125\)

For geese: \(0.96^2 = \frac{\Sigma x_g^2}{18} - 5.2^2 \Rightarrow \Sigma x_g^2 = 503.3088\)

Now, find the combined standard deviation:

\(\text{New } s^2 = \frac{472.6125 + 503.3088}{27} - 5.83^2 = 2.117\)

\(\text{New } s = \sqrt{2.117} = 1.46 \text{ kg}\)

To find the mean, sum all the data points and divide by the number of data points:

\(\text{Mean} = \frac{5 + (-2) + 12 + 7 + (-3) + 2 + (-6) + 4 + 0 + 8}{10} = \frac{27}{10} = 2.7\)

\(To find the variance, use the formula:\)

\(\text{Variance} = \frac{(5^2 + (-2)^2 + 12^2 + 7^2 + (-3)^2 + 2^2 + (-6)^2 + 4^2 + 0^2 + 8^2)}{10} - (\text{Mean})^2\)

\(= \frac{35.1}{10} - 2.7^2 = 3.51 - 7.29 = 27.8\)

(i) To find the mean height of the remaining group of 27 people, we use the formula for the mean:

\(\text{New mean} = \frac{172.6 \times 28 - 161.8}{27} = 173 \text{ cm}\)

(ii) First, find \(\Sigma x^2\) for the original group of 28 people using the formula:

\(\Sigma x^2 = (4.58^2 + 172.6^2) \times 28 = 834728.6\)

Then, for the remaining group of 27 people:

\(\text{Remaining } \Sigma x^2 = 834728.6 - 161.8^2 = 808549.36\)

Finally, calculate the standard deviation for the remaining group:

\(\text{sd of remaining} = \sqrt{\frac{808549.36}{27} - 173^2} = 4.16 \text{ cm}\)

(i) To find the mean weight of the 45 fruits, calculate the total weight of the oranges and lemons and divide by the total number of fruits:

\(\text{Mean} = \frac{(220 \times 20) + (118 \times 25)}{45} = \frac{4400 + 2950}{45} = \frac{7350}{45} = 163\)

(ii) First, find \(\Sigma x_o^2\) and \(\Sigma x_l^2\) using the variance formula:

For oranges:

\(\frac{\Sigma x_o^2}{20} - 220^2 = 32^2\)

\(\Sigma x_o^2 = 20 \times (32^2 + 220^2) = 988480\)

For lemons:

\(\frac{\Sigma x_l^2}{25} - 118^2 = 12^2\)

\(\Sigma x_l^2 = 25 \times (12^2 + 118^2) = 351700\)

\(Combine these to find the variance of all fruits:\)

\(\Sigma x_o^2 + \Sigma x_l^2 = 988480 + 351700 = 1340180\)

\(\text{Variance} = \frac{1340180}{45} - \left(\frac{7350}{45}\right)^2 = 3100 \text{ to } 3120\)

First, calculate the midpoints of each class interval:

• 101 – 120: midpoint = 110.5
• 121 – 130: midpoint = 125.5
• 131 – 135: midpoint = 133
• 136 – 145: midpoint = 140.5
• 146 – 160: midpoint = 153

Calculate the mean using the formula:

\(\bar{x} = \frac{\sum (f \times x)}{\sum f} = \frac{18 \times 110.5 + 48 \times 125.5 + 34 \times 133 + 32 \times 140.5 + 18 \times 153}{150}\)

\(= \frac{1989 + 6024 + 4522 + 4496 + 2754}{150} = 131.9\)

\(Calculate the variance using the formula:\)

\(s^2 = \frac{\sum (f \times x^2)}{\sum f} - (\bar{x})^2\)

\(= \frac{18 \times 110.5^2 + 48 \times 125.5^2 + 34 \times 133^2 + 32 \times 140.5^2 + 18 \times 153^2}{150} - (131.9)^2\)

\(= \frac{3200 + 41400 + 194400 + 157300 + 153600}{150} - 17397.61\)

\(= 137.54\)

\(Standard deviation \(s = \sqrt{137.54} = 11.7 \]\)

(i) For Ashfaq, a suitable statistical diagram is a stem and leaf diagram. This is because it effectively shows the shape, spread, and range of the data, and is particularly useful for a small number of prices, such as the 21 prices collected by Ashfaq. It also allows for easy calculation of the median.

(ii) For Kuljit, a suitable statistical diagram is a histogram. This is because it shows the shape, spread, and modal class of the data, and is suitable for a large number of prices, such as the 163 prices collected by Kuljit. A histogram is more appropriate than a stem and leaf diagram for such a large dataset.

To find the mean, use the formula:

\(\bar{x} = \frac{\sum x}{n}\)

\(\bar{x} = \frac{48.3 + 55.2 + 59.9 + 67.7 + 60.5 + 75.6 + 62.5 + 57.4 + 53.4 + 49.2 + 64.1}{11} = 59.4\)

To find the standard deviation, use the formula:

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

Calculate each squared deviation:

\((48.3 - 59.4)^2, (55.2 - 59.4)^2, (59.9 - 59.4)^2, \ldots, (64.1 - 59.4)^2\)

Sum these squared deviations:

\(\sum (x - \bar{x})^2 = 646.64\)

Then, \(\sigma = \sqrt{\frac{646.64}{11}} = 7.68\)

(i) To find the mean age of all 24 cars, we calculate the total sum of ages and divide by the total number of cars:

For Red Street Garage: \(\text{Total sum} = 3.6 \times 9 = 32.4\)

For Fairwheel Garage: \(\Sigma x = 64\)

Total sum of ages = \(32.4 + 64 = 96.4\)

\(Total number of cars = 9 + 15 = 24\)

Mean age = \(\frac{96.4}{24} = 4.02 \text{ years}\)

\((ii) To find the standard deviation of all 24 cars, we first find the variance:\)

Variance for Red Street Garage: \(\frac{\Sigma x^2}{9} - 3.6^2 = 1.925^2\)

\(\Sigma x^2 = 150\)

Total \(\Sigma x^2 = 150 + 352 = 502\)

\(Mean age = 4.02\)

Variance = \(\frac{502}{24} - 4.02^2 = 4.780\)

Standard deviation = \(\sqrt{4.780} = 2.19 \text{ years}\)

(i) To find the mean \(\bar{x}\), sum the times and divide by the number of students:

\(\bar{x} = \frac{15 + 10 + 48 + 10 + 19 + 14 + 16}{7} = \frac{132}{7} = 18.9\)

To find the standard deviation \(sd\), use the formula:

\(sd = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}\)

Calculate each squared deviation:

• \((15 - 18.9)^2 = 15.21\)
• \((10 - 18.9)^2 = 79.21\)
• \((48 - 18.9)^2 = 841.21\)
• \((10 - 18.9)^2 = 79.21\)
• \((19 - 18.9)^2 = 0.01\)
• \((14 - 18.9)^2 = 24.01\)
• \((16 - 18.9)^2 = 8.41\)

Sum of squared deviations: \(15.21 + 79.21 + 841.21 + 79.21 + 0.01 + 24.01 + 8.41 = 1057.27\)

\(sd = \sqrt{\frac{1057.27}{7}} = \sqrt{151.04} = 12.3\)

(ii) The median is the most appropriate measure of central tendency because it is not affected by the outlier (48).

(iii) The mode is inappropriate because it is 10, which is the lowest value and not representative of the data. The mean is inappropriate because it is affected by the outlier (48), which skews the average.

(i) Since the standard deviation is $0, all rides must cost the same, which is the mean of $2.50. Therefore, the roller coaster and the water slide each cost $2.50 per ride.

(ii) Let the cost of the revolving drum be $x. The equation for the mean cost is:

\(1 \times 2.5 + 3 \times 2.5 + 6 \times x = 3.76 \times 10\)

\(6x = 37.6 - 10\)

\(x = 4.6\)

\(The cost of the revolving drum is $4.60. Now, calculate the variance:\)

\(\sigma^2 = \frac{(2.5^2 \times 1 + 2.5^2 \times 3 + 4.6^2 \times 6)}{10} - 3.76^2\)

\(\sigma = 1.03\)

To find the mean, sum all the lengths and divide by the number of lengths:

\(\bar{x} = \frac{32 + 35 + 45 + 37 + 38 + 44 + 33 + 39 + 36 + 45}{10} = \frac{384}{10} = 38.4 \text{ mm}\)

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

\(s^2 = \frac{1}{10} \left( (32 - 38.4)^2 + (35 - 38.4)^2 + (45 - 38.4)^2 + (37 - 38.4)^2 + (38 - 38.4)^2 + (44 - 38.4)^2 + (33 - 38.4)^2 + (39 - 38.4)^2 + (36 - 38.4)^2 + (45 - 38.4)^2 \right)\)

\(s^2 = \frac{1}{10} (40.96 + 11.56 + 43.56 + 1.96 + 0.16 + 31.36 + 29.16 + 0.36 + 5.76 + 43.56)\)

\(s^2 = \frac{1}{10} (208.4) = 20.84\)

Then, the standard deviation is:

\(s = \sqrt{20.84} \approx 4.57 \text{ mm}\)

(i) In the first round, 32 teams play, resulting in 16 matches. Therefore, 16 teams lose and play only 1 match.

(ii) In the second round, 16 teams play, resulting in 8 matches. Therefore, 8 teams lose and play exactly 2 matches.

(iii) The frequency table is constructed as follows:

(iv) To calculate the mean and variance:
Total matches played = 16(1) + 8(2) + 4(3) + 2(4) + 2(5) = 62
Mean = \(\frac{62}{32} = 1.9375 \approx 1.94\)
For variance, calculate \(\sum f x^2 = 16(1^2) + 8(2^2) + 4(3^2) + 2(4^2) + 2(5^2) = 166\)
Variance = \(\frac{166}{32} - \left(\frac{62}{32}\right)^2 = 1.43\)

To find the median, first arrange the data in ascending order: 40, 40, 41, 42, 45, 47, 48, 49, 50, 51, 352.

The median is the middle value, which is the 6th value in this ordered list: 47.

The median is $47,000.

The mean is not suitable because the data contains an outlier (352), which skews the mean. The mode is not suitable because it does not represent the central tendency effectively in this context.

(i) To find the mean age of the whole group, use the formula for the combined mean:

\(\bar{x} = \frac{(\bar{x}_w \times 10) + (\bar{x}_m \times 13)}{23} = \frac{(41.2 \times 10) + (46.3 \times 13)}{23} = 44.1\)

(ii) First, find \(\Sigma x_w^2\) and \(\Sigma x_m^2\) using the given standard deviations:

\(15.1^2 = \frac{\Sigma x_w^2}{10} - 41.2^2\)

\(\Sigma x_w^2 = 19254.5\)

\(12.7^2 = \frac{\Sigma x_m^2}{13} - 46.3^2\)

\(\Sigma x_m^2 = 29964.74\)

Calculate the total sum of squares:

\(\Sigma = 19254.5 + 29964.74 = 49219.24\)

Finally, find the standard deviation for the whole group:

\(\text{sd} = \sqrt{\frac{49219.24}{23} - 44.1^2} = 14.0\)

To illustrate the data diagrammatically, follow these steps:

1. Choose either pie charts or bar charts for the representation.
2. If using pie charts, create two separate pie charts: one for males and one for females. Divide each pie chart into three sectors representing 'Young', 'Middle-aged', and 'Elderly' with angles proportional to the percentages given (e.g., for males: Young = 40%, Middle-aged = 35%, Elderly = 25%).
3. If using bar charts, create two sets of bars: one set for males and one set for females. Each set should have three bars representing 'Young', 'Middle-aged', and 'Elderly'. Ensure the height of each bar is proportional to the percentage given.
4. Label each chart or set of bars with the appropriate categories and percentages. Include labels for 'Males' and 'Females', as well as 'Young', 'Middle-aged', and 'Elderly'.

To estimate the mean number of hours of sunshine, we first find the midpoints of each class interval:

• For \(30 \leq x < 60\), midpoint = 45
• For \(60 \leq x < 90\), midpoint = 75
• For \(90 \leq x < 110\), midpoint = 100
• For \(110 \leq x < 140\), midpoint = 125
• For \(140 \leq x < 180\), midpoint = 160
• For \(180 \leq x < 240\), midpoint = 210

Next, we calculate the weighted sum of the midpoints:

\(4 \times 45 + 8 \times 75 + 14 \times 100 + 25 \times 125 + 7 \times 160 + 2 \times 210 = 6845\)

The total number of years is 60. Therefore, the estimated mean is:

\(\text{Mean} = \frac{6845}{60} = 114.08\overline{3}\)