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RMIT Classification: Trusted STUDENT STUDENT – KNOWLEDGE ASSESSMENT TASK Task Number 1B of 3 Task Name Assignment PART B National unit/s code MEM30012A National unit/s title Apply Mathematical Techniques in a Manufacturing Engineering or Related Environment National qualification code MEM60112/ 22479VIC National qualification title Advanced Diploma of Engineering (Mechanical) Advanced Diploma of Engineering (Mechanical / Aeronautical) Advanced Diploma of Engineering Technology (Civil Engineering Design) MEM60112 RMIT Program code C6130 / C6131/ C6162 RMIT Course code MATH5268C Section A – Assessment Information Assessment duration and/or due date Week 16-17, Refer Canvas for Exact Date. Task instructions Summary and Purpose of Assessment This assignment in the unit of competency covers the application of the concepts of mathematics to appropriate and simple engineering situations within the individual's area of engineering expertise. You must demonstrate an understanding of: Use two-dimensional geometry to solve practical problems Use trigonometry to solve practical problems Perform basic statistical calculations What and where: This is an individual knowledge-based (problem solving) assessment on Apply Mathematical Techniques in a Manufacturing Engineering or Related Environment. You need to complete this assessment task before its due date. This assessment will take place outside your class time. How: You will be assessed according to the criteria outlined in the assessment below. Please see below “Additional Instructions to Students”. [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 1 of 21
RMIT Classification: Trusted STUDENT Conditions for assessment 1. Students must submit an electronic copy of this assignment via RMIT CANVAS before its due date. 2. Students found in breach of test conditions can be charged with academic misconduct, have their results cancelled, be excluded from the program and receive other penalties. Penalties can also apply if a student’s test material is copied by others. 3. Plagiarism is the presentation of the work, idea or creation of another person as though it is one’s own. It is a form of cheating and is a very serious academic offence that may lead to expulsion from the University. Plagiarised material can be drawn from, and presented in, written, graphic and visual form, including electronic data, and oral presentations. Plagiarism occurs when the origin of the material used is not appropriately cited. 4. RMIT special consideration is to enable you to maintain your academic progress despite adverse circumstances. The process for special consideration can be found at https://www.rmit.edu.au/students/student-essentials/assessment-and-exams/assessment/special- consideration 5. Students with a disability or long-term medical or mental health condition can apply for adjustments to their study and assessment conditions (Reasonable Adjustments and Equitable Assessment Arrangements) by registering with Equitable Learning Services (ELS) at https://www.rmit.edu.au/students/support-and-facilities/student-support/equitable-learning-services Additional Instructions: 1. Performance requirement: a. Must attempt all the questions b. Satisfactory (S) performance successfully complete at least 1 question correctly from each group. There are all together 9 Groups (A to I). c. Not Yet Satisfactory (NYS) performance unable to successfully complete at least 1 question correctly in each group. 2. You are eligible to resubmit the assessment one more time within allowed time frame, if your first attempt result is NYS. 3. Result from this assessment will count towards the final result of competent only if the result is satisfactory (S) 4. Late submission approval will be in line with RMIT policy Application for extension of submittable work (7 calendar days or less) form. 5. Special consideration for assessment Application for special consideration form. 6. If a student’s result is NYS: feedback and agreed deadline date for the student to demonstrate competency will be advised. 7. The date(s) must be within enrolment dates for this competency otherwise the student will be given a final result of NYS and the student may need to re-enrol/ repeat this course. Instructions on submitting Knowledge Assessment You are required to submit this assessment task in CANVAS assignment submission page before its due date. Equipment/resources students must supply (if applicable): Equipment/resources to be provided by RMIT or the workplace (if applicable): Writing pens/pencil, ruler and the like Scientific calculator or similar Computer and RMIT Internet Access RMIT classrooms Microsoft Office Suite [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 2 of 21
RMIT Classification: Trusted STUDENT Section B – Marking Guide: Two Dimensional Geometry Group A Q1: Express the angle “ θ ” in radian between the horizontal position and the wings of the airplane? (PC 3.1) Satisfact ory response Ye s N o A: Angle between the two axes is 90°. Given angle is 63°. Find θ. Subtract given angle from 90°. 90° - 63° = 27° Converting angle from degrees to radians using conversion factor 1° = π 180 27° = π x27 180 Simplify: = 3π 20 = 3 x 3.14 20 = 0.417 radians. Q2: Convert 0.749 radians into equivalent nearest degree. (PC 3.1) Satisfact ory [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 3 of 21
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RMIT Classification: Trusted STUDENT response Ye s N o A: 0.749 x 180 ° Π = 0.749 x 180 ° 3.14 = 134.82 ° 3.14 = 42.9° To the Nearest degree: = 43° Group B Q3: A piece of metal has been cut in the shape of a sector of a circle as shown below. What is the area of this piece of metal? (PC 3.2) Satisfactor y response Yes No A: Converting 120° into radians. = 120 x π 180 = 2.0947 Area of the piece of metal = A = ½ r 2 θ = ½ x 12 2 x 2.0947 = ½ x 144 x 2.0947 = 301.637 [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 4 of 21
RMIT Classification: Trusted STUDENT 2 = 150.8 cm 2 Q4: A structural supporting member is made in the shape shown in below. What is the outside perimeter and shaded area of the structural member? (PC 3.2) Satisfactor y response Yes No A: P = 4 x a P = 4 x 4” P = 18”in Area of shaded region = 5” x 4” = 20” 4.75” x 3.75” = 17.8125 = 20 – 17.8125 = 2.1875 in 2 Group C Q5: The liquid oxygen tank in the external tank of the space shuttle resembles a combination of a hemisphere, a cylinder, and a cone. Find the volume of the liquid oxygen tank. (PC 3.3) Satisfactor y response Yes No [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 5 of 21
RMIT Classification: Trusted STUDENT A: Volume of hemisphere = 2/3 x π x r 3 = 2/3 x 3.14 x 4.2 3 = 2/3 x 3.14 x 74.088 = 2/3 x 232.64 = 465.27/3 = 155.17 Volume of Cylinder = π x r 2 x h = 3.14 x 4.2 2 x 4 = 3.14 x 17.64 x 4 = 3.14 x 70.56 = 221.67 Volume of Cone = 1/3 x π x r 2 x h = 1/3 x 3.14 x 17.64 x 8.1 = 1/3 x 3.14 x 142.884 = 1/3 x 448.656 = 149.63 Sum of all volumes = 155.17 + 221.67 + 149.63 = 526.47m 3 Q6: Find the total surface area of (exterior and interior) of the PVC pipe shown. (PC 3.3) Satisfactor y response Yes No A: Exterior surface area: = 2 x π x r 2 + 2 x π x r x h = 2 x π x (5) 2 + 2 x π x (5)(3) = 2 x π x (25) + 2 x π x (15) = 50 x π + 30 x π = 80π in 2. 8.94 x 3.142 = 28.09. 251.36 Interior Surface area: = 2 π x r 2 + 2 x π x r x h = 2 π x (1.6) 2 + 2 x π x (1.6) (3) = 2 π x (2.56) + 2 x π x (4.8) [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 6 of 21
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RMIT Classification: Trusted STUDENT =5.12 π + 9.6π = 14.72π. 46.25 Add the exterior and interior surface areas to find the total surface area: 80 π in 2 + 14.72 π in 2 = 94.72 π in 2 Group D Q7: The figure below shows five equally spaced holes are drilled on an 80 mm pitch circle diameter metal plate. Calculate their co-ordinates relative to axes 0 x and 0 y in (i) Polar form, (ii) Cartesian form. (PC 3.4) Satisfactor y response Yes No A: Given: Diameter of the circle = 80mm Radius = r = 40mm The hole 2 lies on the y-axis Its argument, θ = π/2 Its polar co-ordinates are (r,θ) = (40,π/2) Angle between any two consecutive holes = 2 π/5 Argument of hole 1 = π/2 – 2π/2 = π/10 [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 7 of 21
RMIT Classification: Trusted STUDENT Argument of hole 3 = π/2 + 2π/5 = 9π/10 Argument of hole 4 = 2π/10 + 2π/5 = 13π/10 Argument of hole 5 = 13π/10 + 2π/5 = 17π/10 The polar co-ordinates of hole 1= (40, π/10) hole 2= (40, π/2) hole 3= (40, 9π/10) hole 4= (40, 13π/10) hole 5= (40, 17π/10) cartesian coordinates; if polar coordinates are (r, θ) then the cartesian coordinates are (x, y) = (r cos θ, r sin θ) the cartesian coordinates of hole 1: = {40cos ( π/10), 40sin(π/10)} = {40(0.9510565), 40 (0.30901070)} = (38.042, 12.361). The cartesian coordinates of hole 2: = {40cos (π/2), 40sin (π/2)} = {40(0), 40(1)} = (0, 40). The cartesian coordinates of hole 3: = {40cos (9 π/10), 40sin (9 π/10)} = {40(-0.9510565), 40(0.3090170)} = ( -38.042, 12.361). The cartesian coordinates of hole 4: = (40cos (13 π/10), 40sin 913 π)} = {40 (0.587785), 40(-0.89017)} = (23.511, -32.381) Q8: (i) Convert (25, 36) Cartesian coordinate into equivalent polar coordinate. Satisfactor y response [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 8 of 21
RMIT Classification: Trusted STUDENT (ii) Express (4.5, 5.16 rad) polar coordinate in Cartesian co- ordinates. (PC 3.4) Yes No A: - To convert Cartesian coordinates (25,36) into polar coordinates, we can use the following formulas: r = √x ² + y ² θ = tan -1 (y/x) Given the Cartesian coordinates (25, 36), we can calculate the polar coordinates as follows: r = 25 2 + 36 2 r = 625+1,296 r = 1,921 r ≈ 43.83 θ=tan -1 (3625) ≈55.22° ≈0.964 Therefore, the equivalent polar coordinates are approximately (43.83,0.964). - To express the polar coordinates (4.5,5.16rad) in Cartesian coordinates. Polar coordinates (r, 𝜃 ) = (4.5,295.65 ° )              Since 5.16 radians = 295.65 degrees Use the following formulas: x=rcos 𝜃 y=rsin 𝜃 The polar coordinates (4.5,295.650), we can calculate the Cartesian coordinates as follows: [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 9 of 21
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RMIT Classification: Trusted STUDENT X = 4.5×cos (295.65 ° ) ≈ 4.5×0.433 ≈ 1.9485 y = 4.5×sin (295.650) ≈ 4.5×−0.9014 ≈ −4.0563 Therefore, the equivalent Cartesian coordinates are approximately (1.97, −4.04). Trigonometry Group E Q9: A car weighing 1500N is on a 20 ° hill. Find the components of the car’s weight parallel “ W x ” and perpendicular “ W y to the road. (PC 4.1) Satisfactor y response Yes No A: Given that: 𝜃 = 20° Explanation: [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 10 of 21
RMIT Classification: Trusted STUDENT Parallel component of weight to the road force Wx = Wsin 𝜃 Perpendicular component of weight to the force Wy = Wcos 𝜃 Wx=1,500×sin20° =0.342×1,500 =513N Wy=1,500×cos20 ° =1,500×0.9397 = 1410N Therefore Wy = 1409.55N and Wx = 513N. Q10: A surveyor is to determine the height of a tower. The transit is positioned at a horizontal distance of 35 meters from the foot of the tower. An angle of elevation of 58 ° is read in sighting the top of the tower. The height from the ground to the transit telescope is 1.70 meters. This is shown in the diagram below. Determine the height “ h ” of the tower. Satisfactor y response Yes No [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 11 of 21
RMIT Classification: Trusted STUDENT (PC 4.1) A: Distance from tower to man = 35m Angle of elevation = 58° Tan 58° = opposite side/adjacent side Tan 58 = x/35 1.60033 = x/35 x = 1.60033(35) x = 56.0117 total height = x +1.7 h = 57.71 total height of tower = h = 57.71m [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 12 of 21
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RMIT Classification: Trusted STUDENT Group F Q11: A highway entrance ramp rises 11.4 m in a horizontal distance of 55 m. What is the measure of the angle in degree that the ramp makes with the horizontal? Give the answer in the nearest whole number. (PC 4.2) Satisfactor y response Yes No A: We know that, tan 𝜃 = Ramp rise/Horizontal Distance   Where, Ramp rise = 11.4 m Horizontal Distance = 55 m tan 𝜃 =11.4/55   𝜃 =11.71° 𝜃 = 12 ° [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 13 of 21
RMIT Classification: Trusted STUDENT Q12: shape of a right triangle measure 25.50 cm and 28.18 cm as shown in the diagram below. What are the measures of the two acute angles “ θ and “ α ” of the piece in degrees? (PC 4.2) Satisfactor y response Yes No A: - Cos 𝜃 = ½ (25.5/28.18) - 𝜃 = ½ cos -1 (25.5/28.18) - 𝜃 = 12° - sin α = 25.5/28.18 - α = sin -1 (25.5/28.18) - α = 64.81° Group G Q13: A crank mechanism of a petrol engine is shown below. Arm OA is 10.0 cm long and rotates clockwise about O . The connecting rod AB is 30.0 cm long and end B is constrained to move horizontally. For the position shown below determine: (i) The angle ABO in degree (ii) The length OB Hint: Use Sine Law. (PC 4.3) Satisfactor y response Yes No A: (i) Given that, OA = 10cm and AB = 30cm [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 14 of 21
RMIT Classification: Trusted STUDENT Using sine law = a/sin A = b/sinB = c/sinC In ABO: Let angleB = 𝜃 10/sin𝜃 = 30/sin50 sin𝜃 = 10/30 x sin50 ° sin𝜃 = 0.255 On taking inverse of sin to find the angle. 𝜃 = sin -1 (0.255) 𝜃 = 14.79 ° = 15° (ii) The sum of all angles in a triangle = 180° AngleA = 180 – (15 + 50) AngleA = 115.21° Using sin law a/sin115.21° = 30/sin50° a = (30/sin50°) x sin 115.21° a = 35.43cm OB = 35.43cm The angle ABO in degree = 15 ° The length OB = 35.4cm [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 15 of 21
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RMIT Classification: Trusted STUDENT Q14: A room 8.0m wide has a span roof which slopes at 33 ° on one side and 40 ° on the other as shown below. Find the length of the roof slope AB . Hint: Use Sine Law. (PC 4.3) Satisfactor y response Yes No A: - Recognize that the sum of the angles in a triangle is 180 ° . Therefore, for the triangle formed by the roof and the ground, we have  𝑚 ∠𝐴 + 𝑚 ∠𝐵 + 𝑚 ∠𝐶 =180 ° Given that  𝑚 ∠𝐴 =33 ° and  𝑚 ∠𝐶 =40 ° calculate  𝑚 ∠𝐵  using the sum of angles in a triangle: 𝑚 ∠𝐵 =180 ° −33 ° −40 ° 𝑚 ∠𝐵 =107 ° Apply the Law of Sines to find the length of the roof slope AB: 𝑏 / sin ∠𝐵 = 𝑐 / sin ∠𝐶 𝐴 𝐶 / sin ∠𝐵 = 𝐴 𝐵 / sin ∠𝐶 8.0/sin107 ° = 𝐴 𝐵 / sin40 ° Solve for AB: 𝐴 𝐵 ⋅ sin107 ° = 8.0 sin40 ° 𝐴 𝐵 =8.0 sin40 °/ sin107 ° Calculate AB: 𝐴 𝐵 ≈5.38 m So, the length of the roof slope AB is approximately 5.4m. [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 16 of 21
RMIT Classification: Trusted STUDENT Group H Q15: A metal frame in the shape of an oblique triangle is to be fabricated. One side of the frame is 2.4 meters long. One end of the second side, which is 1.80 meters long, is to be fastened to an end of the 2.4 meters side at an angle of 58.008 as shown in the diagram below. Compute the required length of the third side of the frame. Hint: use Cosine law . (PC 4.4) Satisfactor y response Yes No A: Identify the given sides of the triangle and the angle between them. The sides are  𝑎 =1.8 m  and  𝑏 =2.4m , and the angle between them is  𝐶 =58.00 Use the Cosine Law to find the length of the third side c, which is not given directly. The Cosine Law states that 𝑐 2 = 𝑎 2 + 𝑏 2 −2 𝑎 𝑏 cos( 𝐶 ) Substitute the given values into the Cosine Law formula: 𝑐 2 = 1.8 2 + 2.4 2 – 2 x 1.8 x 2.4 x cos(58.00 ) Calculate the value of  𝑐 2 : 𝑐 2 =3.24 + 5.76 – 2 x 1.8 x 2.4 x 0.5299 𝑐 2 =9.00 − 4.679088 𝑐 2 = 4.320912 Take the square root of  𝑐 2  to find the length of the third side c: 𝑐 = 4.320912 𝑐 ≈ 2.10m Therefore, the required length of the third side of the frame is approximately 2.10 meters. [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 17 of 21
RMIT Classification: Trusted STUDENT Q16: Find angle B for the roof truss shown in the figure below. Hint: Use Cosine law . (PC 4.4) A: Law of cosine: For any triangle ABC Consider the sides AB = C, BC = A, AC = B The law of cosines defined as, CosA = b 2 + c 2 – a 2 /2bc, CosB = a 2 + c 2 – b 2 / 2ac, CosC = a 2 + b 2 - c 2 /2ab ABC sides – a = 11, b = 6, c = 9. Finding angleB using law of cosine CosB = 11 2 + 9 2 – 6 2 / 2 x 11 x 9 CosB = 121 + 81 + - 36/198 CosB = 0.8384 B = Cos -1 (0.8384) B = 33.03 B = 33 ° Basic statistics Group I Q17: The lengths of forty nails measured to the nearest centimetre are: 4,5,3,7,3,3,5,4,6,8,8,8,6,1,7,3,6,4,6,3 [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 18 of 21
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RMIT Classification: Trusted STUDENT 4,6,8,9,5,6,5,2,3,2,6,4,8,1,5,7,6,3,5,5 Calculate i. Mean ii. Median iii. Mode iv. Standard deviation and v. Interpret the data on the graphical representation with an aid of any computer software (computational analysis) (PC 7.1, 7.2) A: Arranging data in ascending order: 1,1,2,2,3,3,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,7,7,7,8,8,8,8,8,9 Mean = sum of all values divided by number of values: =1+1+2+2+3+3+3+3+3+3+3+4+4+4+4+4+5+5+5+5+5+5+5+6+6+6+6+6+6+6+6+7+7+7+8+8+ 8+8+8+9/40 = 200/40 Mean = 5 Median = middle value of the total given values. Median =5 + 5/2 = 5 Median = 5 Mode = Value that appears the most often Mode = 6 Standard deviation = s = √ sum of individual value – Mean/ total number of values -1 s =√ 164/40-1 s = 2.02 Q18: A manufacturer of printed wring boards used in the assembly of personal computers is testing the quality of its soldering process. To test the strength of the solder joints, some of the soldered leads are pulled until they either break or come off the board. The pull strength measured in pounds for 30 tests are shown below. 59.0 58.0 55.0 63.0 62.0 64.0 62.0 60.0 64.0 60.0 55.0 68.0 66.0 65.0 65.0 64.0 53.0 68.0 66.0 66.0 69.0 59.0 66.0 66.0 68.0 61.0 68.0 62.0 64.0 65.0 By using the given data calculate: [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 19 of 21
RMIT Classification: Trusted STUDENT Calculate vi. Mean vii. Median viii. Mode ix. Standard deviation and x. Interpret the data on the graphical representation with an aid of any computer software (computational analysis) (PC 7.1, 7.2) A: Arranging data in ascending order: 53.0, 55.0, 55.0, 58.0, 59.0, 59.0, 60.0, 60.0, 61.0, 62.0, 62.0, 62.0, 63.0, 64.0, 64.0, 64.0, 64.0, 65.0, 65.0, 65.0, 66.0, 66.0, 66.0, 66.0, 66.0, 68.0, 68.0, 68.0, 68.0, 68.0. Mean = sum of all values divided by number of values: = 53.0 + 55.0 + 55.0 + 58.0 + 59.0 + 59.0 + 60.0 + 60.0 + 61.0 + 62.0 + 62.0 + 62.0 + 63.0 + 64.0 + 64.0 + 64.0 + 64.0 + 65.0 + 65.0 + 65.0 + 66.0 + 66.0 + 66.0 + 66.0 + 66.0 + 68.0 + 68.0 + 68.0 + 68.0 + 68.0/30 Mean = 63.03 Median = middle value of the total given values. Median = 64 + 64/2 = 128/2 Median = 64 Mode = Value that appears the most often Mode = 66 Standard deviation = s = √ sum of individual value – Mean/ total number of values -1 S = 1831 – 63.03/30 – 1 S = 4.11 [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 20 of 21
RMIT Classification: Trusted STUDENT Section C – Feedback to Student Has the student successfully completed the task? Yes No Feedback to student: Assessor Name Date [C6130 MEM30012A (MATH5268C) Student Assessment 2 ] [ 1B of 3] [ 01/07/2022 ] Student knowledge assessment task © Content is subject to copyright, RMIT University FINAL APPROVED – STUDENT KNOWLEDGE ASSESSMENT TASK TEMPLATE – June 2019_Version 2.0 Page 21 of 21
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