University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

459

Pages:

5
Author:

Ruby Bates

Question 1 Find the following for the given data: Arithmetic Mean Median All Quartiles Answer: Step 1: Given data: Units Consumed 1 – 50 51 – 100 101 – 150 151 – 200 201 – 250 251 – 300 Frequency 68 104 126 84 44 --- Step 2: To Find Arithmetic Mean We calculate the arithmetic mean using the assumed mean method: Class Frequency (f) Mid Value (x) 𝑑 = 𝑋 − 𝐴 ℎ d= h X−A 𝑓 ⋅ 𝑑 f⋅d Cumulative Frequency 𝑐 𝑓 cf 1 – 50 68 25.5 -2 -136 68 51 – 100 104 75.5 -1 -104 172 101 – 150 126 125.5 0 0 298 151 – 200 84 175.5 1 84 382 201 – 250 44 225.5 2 88 426 Assumed Mean 𝐴 = 125.5 A=125.5, ℎ = 50 h=50. 𝑋 ˉ = 𝐴 + Σ 𝑓 ⋅ 𝑑 𝑁 ⋅ ℎ X ˉ =A+ N Σf⋅d ⋅h Substituting values: 𝑋 ˉ = 125.5 + − 68 426 ⋅ 50 X ˉ =125.5+ 426 −68 ⋅50 𝑋 ˉ = 125.5 − 7.9812 = 117.5188 X ˉ =125.5−7.9812=117.5188 Arithmetic Mean: 117.52 Step 3: To Find Median Median = 𝐿 + 𝑁 2 − 𝑐 𝑓 𝑓 ⋅ ℎ Median=L+ f 2 N −cf ⋅h Where: 𝐿 = Lower boundary of median class = 100.5 L=Lower boundary of median class=100.5, 𝑁 = 426 N=426, 𝑐 𝑓 = 172 cf=172, 𝑓 = 126 f=126, ℎ = 50 h=50. Median = 100.5 + 213 − 172 126 ⋅ 50 Median=100.5+ 126 213−172 ⋅50 Median = 100.5 + 41 126 ⋅ 50 = 100.5 + 16.2698 = 116.7698 Median=100.5+ 126 41 ⋅50=100.5+16.2698=116.7698 Median: 116.77 Step 4: To Find Quartiles Quartile 1 ( 𝑄 1 Q 1 ) 𝑄 1 = Class with 𝑁 4 𝑡 ℎ observation in cf column Q 1 =Class with 4 N th observation in cf column 𝑁 4 = 426 4 = 106.5 4 N = 4 426 =106.5 The 106. 5 𝑡 ℎ 106.5 th observation lies in the 51 – 100 class. 𝑄 1 = 51 – 100 Q 1 =51–100 Final Results: Arithmetic Mean: 117.52 Median: 116.77 Quartile 1: 51–100 class Question 3 Yorkies Company has fixed costs of $80,000 per month. Variable costs are $12 per item, and the selling price per unit is $20. Determine the loss probability. Answer: Profit Formula: Profit = 20 𝑥 − ( 12 𝑥 + 80000 ) = 8 𝑥 − 80000 Profit=20x−(12x+80000)=8x−80000 Set Profit = 0: 8 𝑥 − 80000 = 0 ⟹ 𝑥 = 10000 8x−80000=0⟹x=10000 Convert Output to Z-Score: Mean 𝜇 = 11100 μ=11100, Standard deviation 𝜎 = 550000 = 741.62 σ= 550000 =741.62. 𝑧 = 10000 − 11100 741.62 ≈ − 1.48 z= 741.62 10000−11100 ≈−1.48 From Z-table, 𝑃 ( 𝑧 < − 1.48 ) = 0.0694 = 6.94 % P(z<−1.48)=0.0694=6.94%. Probability of Loss: 7% Question 4 Construct a frequency distribution table for the data and calculate relative frequencies. Modules Registered Frequency 10 5 9 4 8 3 Frequency Distribution Table: Classes Frequencies Relative Frequency (%) Cumulative Frequency (%) 0 – 3 9 30 30 3 – 6 11 37 67 6 – 9 8 27 93 10 or more 2 7 100 Final Table: Relative Frequency: Above Cumulative Frequency: Above