Range = 5

Largest value = 50

50 - Least = 5

50 - 5 = Least

45 = Least

Mean = 16

Number of numbers = 10

$\frac{\mathrm{Sum\; of\; numbers}}{10}$ = 16

Get rid of the fraction by multiplying both sides by the denominator 10.

Sum of numbers = 16 x 10 = 160

When k is added, the new sum = 160 + k and the number of numbers become 10 + 1 = 11

Mean for the eleven (11) numbers is 18

$\frac{\mathrm{160\; +\; k}}{11}$ = 18

Get rid of the fraction by multiplying both sides by the denominator 11

160 + k = 18 x 11

160 + k = 198

k = 198 - 160

k = 38

Mean = $\frac{\mathrm{Sum\; of\; numbers}}{\mathrm{Number\; of\; numbers}}$

Mean = $\frac{\mathrm{(x\; +\; y)\; +\; (2x\; +\; 3y)\; +\; (2x\; -\; 2y)\; +\; (3x\; -\; 2y)}}{4}$

Mean = $\frac{\mathrm{x\; +\; y\; +\; 2x\; +\; 3y\; +\; 2x\; -\; 2y\; +\; 3x\; -\; 2y}}{4}$

Group like terms

Mean = $\frac{\mathrm{x\; +\; 2x\; +\; 2x\; +\; 3x\; +\; y\; +\; 3y\; -\; 2y\; -\; 2y}}{4}$

Mean = $\frac{\mathrm{8x\; +\; 0}}{4}$

Mean = $\frac{\mathrm{8x}}{4}$ = 2x

Mean = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{150}{30}$ = 5

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

Sum of number of students (frequencies) = 2 + 1 + 4 + 3 + 8 + 5 + 4 + 3

Sum of number of students (frequencies) = 30

Median position = $\frac{\mathrm{\Sigma f\; +\; 1}}{2}$ = $\frac{\mathrm{30\; +\; 1}}{2}$ = $\frac{31}{2}$ = 15.5

∴ the median is the average of the 15^th and the 16^th position

The 15^th and the 16^th positions fall in the number of subjects = 5

Median = $\frac{\mathrm{5\; +\; 5}}{2}$ = $\frac{10}{2}$ = 5

Average = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Number of items = 3 + 9
Number of items = 12

Average cost = $\frac{\mathrm{Sum\; of\; cost}}{12}$

Sum of cost = 3 x 0.4 + 9 x 0.2
Sum of cost = 1.2 + 1.8
Sum of cost = GH₵3

Average cost = $\frac{\mathrm{GH₵3}}{12}$

Average cost = GH₵ 0.25

Range = Largest Value - Smallest Value

Range = 83 - 47

Range = 36

Mean = $\frac{\mathrm{Sum\; of\; numbers}}{\mathrm{Number\; of\; numbers}}$

Mean = $\frac{\mathrm{83\; +\; 47\; +\; 62\; +\; 49\; +\; 55\; +\; 72\; +\; 58\; +\; 62}}{8}$

Mean = $\frac{488}{8}$ = 61

Mean/Average = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Sum of the fractions = (1 + 2 + 3 + 4 + 5)[$\frac{1}{2}$ + $\frac{2}{3}$ + $\frac{3}{4}$ + $\frac{4}{5}$ + $\frac{5}{6}$ ]

The L.C.M of 2, 3, 4, 5 and 6 is 60.

Sum of the fractions = 15[$\frac{\mathrm{30x\; 1\; +\; 20\; x\; 2\; +\; 15\; x\; 3\; +\; 12\; x\; 4\; +\; 10\; x\; 5}}{60}$]

Sum of the fractions = 15[$\frac{\mathrm{30\; +\; 40\; +\; 45\; +\; 48\; +\; 50}}{60}$]

Sum of the fractions = 15[$\frac{213}{60}$]

Sum of the fractions = 15$\frac{213}{60}$ = $\frac{\mathrm{15\; x\; 60\; +\; 213}}{60}$ = $\frac{1113}{60}$

Number of numbers = 5

Mean = $\frac{1113}{60}$ ÷ 5

Every number is being divided by 1 but not written.

5 = $\frac{5}{1}$

Mean = $\frac{1113}{60}$ ÷ $\frac{5}{1}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

Mean = $\frac{1113}{60}$ x $\frac{1}{5}$

Mean = $\frac{\mathrm{1113\; x\; 1}}{\mathrm{60\; x\; 5}}$

Mean = $\frac{1113}{300}$ = 3.71

Arrange the numbers in ascending order.

6, 7, 7, 7, 8, 8, 8, 9, 9, 11

First Quartile (ungrouped data) = $\frac{\mathrm{n}}{4}$^th value

Where n is the number of items.

n = 10

First Quartile (ungrouped data) = $\frac{10}{4}$ = 2.5^th value

First Quartile = $\frac{\mathrm{7\; +\; 7}}{2}$ = 7.0

Mean = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{6030}{37}$ = 162.97297 ≈ 163.0 (one decimal place)

The sum of all the angles is 360°

Angle of sector for orange = 360° - (100° + 80° + 60°)

Angle of sector for orange = 360° - 240°

Angle of sector for orange = 120°

The quantity for apple is known to be 60.

The angle of sector for apple is 80°

If 80° = 60

120° = $\frac{\mathrm{60\; x\; 120°}}{\mathrm{80°}}$ = 90

∴ there are 90 oranges.

360° represents the total population of the city.

Sum of all the angles should be 360°

Angle of sector for men = 360° - (120° + 169°)

Angle of sector for men = 71°

If 360° = 1,800,000

71° = $\frac{\mathrm{1800000\; x\; 71°}}{\mathrm{360°}}$ = 355000

Σ(x_i - µ)² = (15 - 21)² + (21 - 21)² + (17 - 21)² + (26 - 21)² + (18 - 21)² + (29 - 21)²

Σ(x_i - µ)² = 36 + 0 + 16 + 25 + 9 + 64

Σ(x_i - µ)² = 150

N = 6

σ = $\sqrt{\frac{150}{6}}$ = 5

Upper quartile - Lower quartile (25th percentile) = Interquartile range

Lower quartile (25th percentile) = 16

Interquartile range = 7

Upper quartile - 16 = 7
Upper quartile = 7 + 16
Upper quartile = 23

Mean (x) = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean (x) = $\frac{\mathrm{2\; +\; 4\; +\; 7\; +\; 8\; +\; 9}}{5}$ = $\frac{30}{5}$ = 6

Σfx² = 1 x 2² + 1 x 4² + 1 x 7² + 1 x 8² + 1 x 9²

Σfx² = 4 + 16 + 49 + 64 + 81 = 214

Σf = 1 + 1 + 1 + 1 + 1 = 5

Variance = $\frac{214}{5}$ -6² = 6.8

Angle of sector of orange = 360° - (100° + 60° + 120°)

Angle of sector of orange = 360° - 280°

Angle of sector of orange = 80°

Angle of sector of apple = 120°

Number of oranges = 60

If 80° = 60

120° = $\frac{\mathrm{60\; x\; 120°}}{\mathrm{80°}}$ = 90

There are 90 apples

The mode is the one which occurs the most/ the one with the highest frequency/number of students

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

8, 8, 10, 11, 13, 13, 13, 14, 14, 15, 16, 17, 18, 18

There are two numbers 13 and 14 at the middle.

Median = $\frac{\mathrm{13\; +\; 14}}{2}$ = 13.5

Mean = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{8\; +\; 8\; +\; 10\; +\; 11\; +\; 13\; +\; 13\; +\; 13\; +\; 14\; +\; 14\; +\; 15\; +\; 16\; +\; 17\; +\; 18\; +\; 18}}{14}$

Mean = $\frac{188}{14}$ = 13.4286 ≈ 13.4

7 students scored more than 13.4 (14, 14, 15, 16, 17, 18 and 18)

Mean (x̄) = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{240}{15}$ = 16 years

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

Method I

Σf = 5 + 6 + 3 + 1 = 15

Median position = $\frac{\mathrm{\Sigma f\; +\; 1}}{2}$ = $\frac{\mathrm{15\; +\; 1}}{2}$ = $\frac{16}{2}$ = 8 position

5 position for 15 years

5 + 6 = 11 position for 16 years

∴ The 8^th position falls under 16 years

Method II

List the ages and pick the middle age.

Note: you list each age n times. Where n is the No. of students.

15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 18

Notes:

1. Interval for the minutes axis = 2.45 - 1.95 = 0.5

2. Each small grid = $\frac{0.5}{10}$ = 0.05

3. Interquartile range of the distribution = Q₃ − Q₁

Q₃ position = $\frac{\mathrm{3\; x\; \Sigma f}}{4}$

Q₃ position = $\frac{\mathrm{3\; x\; 50}}{4}$ = 37.5 position

Corresponding minutes for 37.5 is 2.95 + 5 x 0.05 = 3.2 minutes

Q₃ = 3.2 minutes

Q₁ position = $\frac{\mathrm{1\; x\; \Sigma f}}{4}$

Q₁ position = $\frac{\mathrm{1\; x\; 50}}{4}$ = 12.5 position

Corresponding minutes for 12.5 is 2.45 minutes

Q₃ = 3.2 minutes

Q₁ = 2.45

Interquartile range of the distribution = 3.2 - 2.45 = 0.75

At least 3 minutes means 3 minutes or more.

Number of customers who waited for at least 3 minutes = 50 - 32 = 18

Notes:

1. Interval for the minutes axis = 2.45 - 1.95 = 0.5

2. Each small grid = $\frac{0.5}{10}$ = 0.05

3. 3 minutes is 1 grid (0.05) after 2.95. 3 = 2.95 + 0.05

4. The corresponding number of customers for the 3 minutes is 32

Mean (x̄) = $\frac{\mathrm{5\; +\; 8\; +\; 3\; +\; 7\; +\; 2}}{5}$ = $\frac{25}{5}$ = 5

|(x - x̄)| for 5 = 5 - 5 = 0

|(x - x̄)| for 8 = 8 - 5 = 3

|(x - x̄)| for 3 = 3 - 5 = |-2| = 2

|(x - x̄)| for 7 = 7 - 5 = 2

|(x - x̄)| for 2 = 2 - 5 = |-3| = 3

f = 1 for each

Σf|(x - x̄)| = 1 x 0 + 1 x 3 + 1 x 2 + 1 x 2 + 1 x 3 = 10

Σf = 1 + 1 + 1 + 1 + 1 = 5

Mean Deviation = $\frac{10}{5}$ = 2

Mean (x̄) = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{38}{16}$ = 2.375 ≈ 2.4 (one decimal place)

The number of students who took the test is the sum of all the frequencies.

Calculate the value of each grid to determine the frequencies.

Grid value = $\frac{\mathrm{Interval}}{\mathrm{Number\; of\; grids\; per\; interval}}$

Grid value = $\frac{5}{20}$ = 0.25

Value of frequency = Number of grids x Grid value

Frequency for mark 1 = 12 x 0.25 = 3

Frequency for mark 2 = 8 x 0.25 = 2

Frequency for mark 3 = 20 x 0.25 = 5

Frequency for mark 4 = 32 x 0.25 = 8

Frequency for mark 5 = 56 x 0.25 = 14

Frequency for mark 6 = 64 x 0.25 = 16

Frequency for mark 7 = 40 x 0.25 = 10

Frequency for mark 8 = 28 x 0.25 = 7

Frequency for mark 9 = 24 x 0.25 = 6

Frequency for mark 10 = 4 x 0.25 = 1

Number of students = 3 + 2 + 5 + 8 + 14 + 16 + 10 + 7 + 6 + 1

Number of students = 72

The number of students who took the test is the sum of all the frequencies.

Calculate the value of each grid to determine the frequencies.

Grid value = $\frac{\mathrm{Interval}}{\mathrm{Number\; of\; grids\; per\; interval}}$

Grid value = $\frac{5}{20}$ = 0.25

Value of frequency = Number of grids x Grid value

Frequency for mark 1 = 12 x 0.25 = 3

Frequency for mark 2 = 8 x 0.25 = 2

Frequency for mark 3 = 20 x 0.25 = 5

Frequency for mark 4 = 32 x 0.25 = 8

Frequency for mark 5 = 56 x 0.25 = 14

Frequency for mark 6 = 64 x 0.25 = 16

Frequency for mark 7 = 40 x 0.25 = 10

Frequency for mark 8 = 28 x 0.25 = 7

Frequency for mark 9 = 24 x 0.25 = 6

Frequency for mark 10 = 4 x 0.25 = 1

Number of students = 3 + 2 + 5 + 8 + 14 + 16 + 10 + 7 + 6 + 1

Number of students = 72

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Number of students who scored below 5 marks = 3 + 2 + 5 + 8 = 18

Percentage who failed = $\frac{18}{72}$ x 100% = 25%

The number of students who took the test is the sum of all the frequencies.

Calculate the value of each grid to determine the frequencies.

Grid value = $\frac{\mathrm{Interval}}{\mathrm{Number\; of\; grids\; per\; interval}}$

Grid value = $\frac{5}{20}$ = 0.25

Value of frequency = Number of grids x Grid value

Frequency for mark 1 = 12 x 0.25 = 3

Frequency for mark 2 = 8 x 0.25 = 2

Frequency for mark 3 = 20 x 0.25 = 5

Frequency for mark 4 = 32 x 0.25 = 8

Frequency for mark 5 = 56 x 0.25 = 14

Frequency for mark 6 = 64 x 0.25 = 16

Frequency for mark 7 = 40 x 0.25 = 10

Frequency for mark 8 = 28 x 0.25 = 7

Frequency for mark 9 = 24 x 0.25 = 6

Frequency for mark 10 = 4 x 0.25 = 1

Number of students = 3 + 2 + 5 + 8 + 14 + 16 + 10 + 7 + 6 + 1

Number of students = 72

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

Median position = $\frac{\mathrm{\Sigma f\; +\; 1}}{2}$ = $\frac{\mathrm{72\; +\; 1}}{2}$ = $\frac{73}{2}$ = 36.5 position

Add the frequencies until you reach 36. The median is the average of the 36^th and 37^th marks.

3 + 2 + 5 + 8 + 14 = 32

The 36^th and 37^th falls on the 6 marks.

Median mark = $\frac{\mathrm{6\; +\; 6}}{2}$ = $\frac{12}{2}$ = 6 marks

Number of students who obtained 6 marks = 16

Total number of students = 72

P(Median mark) = $\frac{16}{72}$ = $\frac{2}{9}$

Note: 8 divides 16, 2 times and 72, 9 times.

Mean = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$

The mean is not greater than 5 means the mean can be less than 5 or equal to 5.

$\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$ ≤ 5

$\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5

Get rid of the fraction by multiplying both sides by the denominator, 4

4 x $\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5 x 4

14 + 2x ≤ 20

2x ≤ 20 - 14

2x ≤ 6

Divide both sides by 2.

x ≤ $\frac{6}{2}$

x ≤ 3