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Total weight = 15 \times 1000 = 15000 kg. New average = 1000 - 5 = 995 kg. Total for 14 bars = 14 \times 995 = 13930 kg. Heaviest = 15000 - 13930 = 1070 kg.

The pie chart represents a total of 360^\circ. Total number of people = 16 (Doctors) + 8 (Engineers) + 24 (Lawyers) + 21 (Information Technologists) + 27 (Teachers) = 96. The portion for doctors is 16 out of 96. Angle for doctors = \(\frac{16}{96} \times 360^\circ = 60^\circ\).

Total number of students = x + 1 + 2x + 2 + 3 = 3x + 6 = 12 Solve for x: 3x + 6 = 12 3x = 6 x = 2 Now substitute x = 2: - 54 kg: 2 students - 56 kg: 1 student - 58 kg: 4 students (2x = 4) - 60 kg: 2 students - 62 kg: 3 students The frequencies are: 2, 1, 4, 2, 3. The highest frequency is 4 (at 58 kg). Thus, Modal weight: 58 kg

The modal class is the class interval with the highest frequency. From the table: Frequency of 1-10: 2 11 - 20: 7 21 - 30: 10 (highest) 31 - 40: 3 41 - 50: 1 Therefore, the modal class is 21-30. The upper-class boundary of a class interval is the upper limit, considering the continuous nature of grouped data (typically 0.5 above the stated upper limit if classes are assumed continuous). For the class 21 - 30, the upper-class boundary is 30.5.

MODE = 3, the most frequent number.

Total ages of four female teachers = 4 x 40 = 160 Total ages of eight male teachers = 8 x 25 = 200 Total ages of all the teachers in the school = 160 + 200 = 360 Average ages of all the teachers = \(\frac{\text{ total ages of all teachers}}{\text{total number of teachers}}\) = \(\frac{360}{12}\) = 30.

Mean = \(\frac{ 13 + 16 + x + 18 + 21 + 2x + 35}{7}\) = 22 103 + 3x = 154 3x = 154 - 103 = 51 x = \(\frac{51}{3}\) = 17

To find the variance from the standard deviation: Given: Standard deviation = 12.34 Variance = (12.34)\(^2\) = 152.2756 Rounding to the nearest whole number: Variance ≈ 152. ​

mean = \(\frac{ 4 + 5 + 6 + 7 + 8}{5}\) = 6 x x - mean (x - mean)\(^2\) 4 4 - 6 = -2 4 5 5 - 6 = -1 1 6 6 - 6 = 0 0 7 7 - 6 = 1 1 8 8 - 6 = 2 4 Variance = \(\frac{10}{5}\) S.D = \(\sqrt{variance}\) = \(\sqrt{2}\)

Mode = 42 42 occurred thrice.

Class Interval Class Mark Frequency (f) fx 5 - 6 5.5 29 5.5 x 29 = 159.5 7 - 8 7.5 40 7.5 x 40 = 300 9 - 10 9.5 38 9.5 x 38 = 361 \(\sum f = 107\) \(\sum fx = 820.5\) Mean = \(\frac{\sum fx}{\sum f} = \frac {820.5}{107}\) = 7.7 (1 d.p)

The first five prime numbers are 2, 3, 5, 7 and 11 M.D = \(\frac {\sum (x - x̄)}{n}\) n = 5 \(x̄ = \frac {\sum x}{n} = \frac {2 + 3 + 5 + 7 + 11}{5} = \frac {28}{5} = 5.6\) M.D = \(\frac {(2 - 5.6) + (3 - 5.6) + (5 - 5.6) + (7 - 5.6) + (11 - 5.6)}{5}\) = \(\frac {(-3.6) + (-2.6) + (-0.6) + (1.4) + (5.4)}{5} = \frac {13.6}{5}\) Take the absolute value of all numbers in bracket = \(\frac {3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac {13.6}{5}\) \(\therefore M.D = 2.72\)

\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54) Median = \(1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})\)c \(l_m\)=lower class boundary of the median age = 6.5 \(\sum f\) = 107 \(f_b\) = sum of all frequencies before the median age = 29 \(f_m\)=frequency of the median age = 40 c = class width = 8.5 - 6.5 = 2 Median = 6.5 + \((\frac {^{107}/_2 - 29}{40})\) x 2 = 6.5 + \((\frac {53.5 - 29}{40})\) x 2 = 6.5 + \((\frac {24.5}{40})\) x 2 = 6.5 + 1.225 \(\therefore\) median age = 7.725

Number of students who scored atleast 25% = 5 + 3 + 8 = 16

Total number of students = total frequency = 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53 Since the failed mark was 4, Total number of students that passed = 12 + 7 + 5 + 10 = 34 Pr( passed) = \(\frac{34}{53}\) = 0. 64