Stat individual Assignment
.docx
School
Jacob Hespeler Secondary School**We aren't endorsed by this school
Course
MATHEMATICS MCV4U
Subject
Statistics
Date
Dec 31, 2024
Pages
5
Uploaded by MinisterMusicCaribou20
1. The following data are the record of weights in kg of wheat production from 40 farmers.60 64 98 55 64 89 87 65 62 88 67 70 77 69 78 39 75 56 71 5199 68 95 86 57 53 47 50 55 81 80 98 51 46 63 66 85 79 83 70Based on the above dataA.Determine the class, class boundary, and frequency count of the data.B.Determine Cumulative frequency less than type and Cumulative frequency morethan typeC.Present your data using histogram diagram=> To determine the class intervals, we first need to find the range of the data:Range = Highest value - Lowest valueRange = 99 - 39Range = 60Choose the Number of Classes: Using Sturges’ formula: ( k = 1 + 3.322 log N ), where N = 40 :( k = 1 + 3.322 log 40 =approx 7.32). We’ll use 8 classes.=> Next, we need to determine the number of classes. A common rule of thumb is to use between 5and 20 classes. Let's use 5 classes for this data.Class width = Range / number of classesClass width = 60 / 8Class width = 7.5= 8 (rounded)=> Now we can determine the class intervals:1. 39-46. 4. 63-70. 7. 87-942. 47-54. 5. 71-78. 8. 95-1023. 55-62 6. 79-86=> Next, we calculate the class boundaries:1. 38.5-46.5. 4. 62.5-70.5. 7. 86.5-94.52. 46.5-54.5. 5. 70.5-78.5. 8. 94.5-102.53. 54.5-62.5 6. 78.5-86.5=> Next, we determine the frequency count:1. 3 4. 8 7. 4
2. 4 5. 7 8. 43. 5 6. 5=> Cumulative frequency less than type:1. 3 4. 20 7. 362. 7 5. 27 8. 403. 12 6. 32=> Cumulative frequency more than type:1. 40 4. 28 7. 82. 37 5. 20 8. 43. 33 6. 13| Class Boundaries | Frequency | Cumulative (Less Than) | Cumulative (More Than) ||--------------|------------------|-----------|-------------------------|-------------------------|| 39–46 | 38.5–46.5 | 3 | 3 | 40 || 47–54 | 46.5–54.5 | 4 | 7 | 37 || 55–62 | 54.5–62.5 | 5 | 12 | 33 || 63–70 | 62.5–70.5 | 8 | 20 | 28 || 71–78 | 70.5–78.5 | 7 | 27 | 20 || 79–86 | 78.5–86.5 | 5 | 32 | 13 || 87–94 | 86.5–94.5 | 4 | 36 | 8 |
| 95–102 | 94.5–102.5 | 4 | 40 | 4 |=> Finally, we can present the data using a histogram diagram.2. Find the mean, mode, and medianof the data given below. The data representthe population of licensed small business owners in Nifas Silk Lafto sub city.104 104 104 104 104 107 109 109 109 110 109 111 112 111 109To find the mean, mode, and median of the data, we first need to organize the data inascending order: 104 104 104 104 104 107 109 109 109 109 109 110 111 111 112=> Mean: Mean = (Sum of all values) / (Number of values)Mean = (104 + 104 + 104 + 104 + 104 + 107 + 109 + 109 + 109 + 109 + 109 + 110 + 111 + 111+ 112) / 15Mean = 1574 / 15Mean = 104.93 (approx)=> Mode:The mode is the value that appears most frequently in the data set. In this case, the mode is109, as it appears 5 times, which is more frequent than any other value.=> Median:To find the median, we need to first rearrange the data in ascending order:104 104 104 104 104 107 109 109 109 109 109 110 111 111 112Since there are 15 values, the median will be the average of the 8th and 9th values:Median = (109 + 109) / 2 = 109Therefore, the mean of the data is approximately 104.93, the mode is 109, and the median is109.3. A testing lab wishes to test two experimental brands of outdoor paint to see how longeach will last before fading. The testing lab makes 6 gallons of each paint to test. Sincedifferent chemical agents are added to each group and only six cans are involved, these twogroups constitute two small populations. The results (in months) are shown.A.Find the mean of each group (A and B)
B.Find the range of the given dataC.Find the variance and standard deviation for each data set for brand A & B paintData for Brand A:10, 60, 50, 30, 40, 20Data for Brand B: 35, 45, 30, 35, 40, 25A.Brand A: Mean = (10 + 60 + 50 + 30 + 40 + 20)/6 = 210/6 = 35 monthsBrand B: Mean = (35 + 45 + 30 + 35 + 40 + 25)/6 = 210/6 = 35 monthsB.Range of Brand A: 60 – 10 = 50 monthsRange of Brand B: 45 – 25 = 20 monthsC.Variance and Standard Deviation for Brand A:Variance = [(10-35)^2 + (60-35)^2 + (50-35)^2 + (30-35)^2 + (40-35)^2 + (20-35)^2]/6Variance = [(625 + 625 + 225 + 25 + 25 + 225)]/6Variance = 1750/6 = 291.67Standard Deviation = √Variance = √291.67 = 17.08 monthsVariance and Standard Deviation for Brand B:Variance = [(35-35)^2 + (45-35)^2 + (30-35)^2 + (35-35)^2 + (40-35)^2 + (25-35)^2]/6Variance = [(0^2 + 100 + 25 + 0^2 + 25 + 100)]/6Variance = 250/6 = 41.67Standard Deviation = √Variance = √41.67 = 6.46 months