10.Suppose data collected on lifetimes (in hours)of a sample of 390 transistors were tabulated in a frequency distribution and the following results were obtained. Frequency:6 2548 72 116 6038 223 CM_(1)=110; CM_(2)=120 . Where; CM; is class marks of ith class. Assume equal size of the class intervals. Determine: a) The class interval size, class boundaries and class marks of each class. b) Compute the mean, mediar and the modal time of the distribution?

Let's solve the problem step by step.### Part (a): Determine the class interval size, class boundaries, and class marks of each class.Given:- Frequencies: 6, 2548, 72, 116, 6038, 223- Class Marks (CM) for the first two classes: , #### Step 1: Find the class interval sizeThe class mark is the midpoint of a class interval. The difference between consecutive class marks gives us the class interval size. So, the class interval size is 10 hours.#### Step 2: Determine the class boundaries and class marksSince we know the class interval size and the class marks, we can determine the class boundaries and class marks for each class.- **Class 1**: - Class Mark: 110 - Lower Boundary: - Upper Boundary: - Class Interval: [105, 115)- **Class 2**: - Class Mark: 120 - Lower Boundary: - Upper Boundary: - Class Interval: [115, 125)Following this pattern, we can find the class intervals for all classes:- **Class 3**: - Class Mark: 130 - Class Interval: [125, 135)- **Class 4**: - Class Mark: 140 - Class Interval: [135, 145)- **Class 5**: - Class Mark: 150 - Class Interval: [145, 155)- **Class 6**: - Class Mark: 160 - Class Interval: [155, 165)### Part (b): Compute the mean, median, and modal time of the distribution#### Step 1: Compute the MeanThe mean ( ) of a frequency distribution is given by: Where is the frequency and is the class mark.Using the given frequencies and class marks: Total frequency ( ): Mean: #### Step 2: Compute the MedianTo find the median, we need to locate the median class. The median class is the class where the cumulative frequency reaches or exceeds half of the total frequency.Total frequency ( ) = 9003Half of the total frequency ( ) = 4501.5Cumulative frequencies:- Class 1: 6- Class 2: 2554 (6 + 2548)- Class 3: 2626 (2554 + 72)- Class 4: 2742 (2626 + 116)- Class 5: 8780 (2742 + 6038) → Median class- Class 6: 9003 (8780 + 223)The median class is [145, 155) with a cumulative frequency just before it of 2742.Median formula: Where:- = lower boundary of the median class = 145- = total frequency = 9003- = cumulative frequency before the median class = 2742- = frequency of the median class = 6038- = class interval size = 10 #### Step 3: Compute the ModeThe mode is the value that appears most frequently in the data set. For grouped data, the modal class is the class with the highest frequency.The modal class is [145, 155) with a frequency of 6038.Mode formula: Where:- = lower boundary of the modal class = 145- = frequency of the modal class = 6038- = frequency of the class before the modal class = 116- = frequency of the class after the modal class = 223- = class interval size = 10 ### Summary:- **Class Interval Size**: 10 hours- **Class Boundaries and Class Marks**: - Class 1: [105, 115), CM = 110 - Class 2: [115, 125), CM = 120 - Class 3: [125, 135), CM = 130 - Class 4: [135, 145), CM = 140 - Class 5: [145, 155), CM = 150 - Class 6: [155, 165), CM = 160- **Mean**: 141.63 hours- **Median**: 147.91 hours- **Mode**: 150.04 hours