Samacheer Kalvi Class 12 Business Maths Solutions Chapter 9 Applied Statistics Exercise 9.3

 

Question 1. Define Statistical Quality Control.
Answer: Statistical Quality Control (SQC) uses statistical methods to check and improve the quality of products. Quality means the standard of a product, which depends on materials, manpower, machines, and management (often called 4M's). Quality Control makes sure products meet specific standards from when raw materials arrive, through every step of making them, until they are delivered. It helps find any quality issues early on. This continuous check ensures that what is produced meets expectations.
In simple words: Statistical Quality Control uses math to make sure products are made well. It checks everything from raw materials to the final product to catch problems early.

🎯 Exam Tip: Remember to mention the "4M's" (Material, Manpower, Machines, Management) when defining Quality Control, as it's a key concept.

 

Question 2. Mention the types of causes for variation in a production process.
Answer: There are two main types of reasons why a product's quality might vary during production:
1. Chance Causes (also called Random causes)
2. Assignable Causes
These two categories help us understand and manage quality changes.
In simple words: Product quality can change because of two things: normal, random reasons or specific, traceable problems.

🎯 Exam Tip: Listing both types clearly is important. A brief explanation of each in subsequent questions will show deeper understanding.

 

Question 3. Define Chance Cause.
Answer: Chance causes are small, natural changes that are part of any manufacturing process. These variations are beyond human control and cannot be stopped or removed. They do not greatly affect the product's quality. Examples include slight changes due to rain, floods, or power cuts. These are like everyday small changes that happen naturally.
In simple words: Chance causes are small, normal changes in making things that we cannot control. They don't hurt the quality much.

🎯 Exam Tip: Emphasize that chance causes are "natural and inherent" and "beyond human control" to differentiate them from assignable causes.

 

Question 4. Define Assignable cause.
Answer: Assignable causes are specific, non-random reasons for variation in production quality. These causes can happen at any stage, from when raw materials arrive to when the finished product is delivered. Unlike chance causes, assignable causes can be identified and fixed. Important factors contributing to assignable causes include bad raw materials, faulty machines, workers who are not skilled, worn-out tools, or a new production process that isn't working well. These are problems that can be traced back to a specific source.
In simple words: Assignable causes are clear problems that make product quality change. These problems can be found and fixed, like bad tools or untrained workers.

🎯 Exam Tip: When defining assignable causes, mention that they are "non-random" and "can be identified and rectified" as key distinctions from chance causes.

 

Question 5. What do you mean by product control?
Answer: Product control is about managing the quality of a product by carefully checking it using sampling plans. The goal of product control is to guarantee customers a certain level of quality. It tries to make sure that the products sold do not have too many defects. This means it involves sorting raw materials, half-finished goods, or finished goods into what is acceptable and what should be rejected. This is the final check before a product goes to market.
In simple words: Product control checks finished items to make sure they are good enough to sell. It sorts products to remove bad ones.

🎯 Exam Tip: Highlight that product control focuses on the "quality of the product" itself, often through "sampling inspection plans" and classifying items as "acceptable or rejectable".

 

Question 6. What do you mean by process control?
Answer: Process control focuses on reducing the number of faulty items made during the production process. This is achieved by using control charts, which help monitor the process as it happens. By watching the process, problems can be found and fixed before many defective products are made. It's about ensuring the making process itself is stable.
In simple words: Process control tries to reduce bad items by checking how things are made. It uses control charts to keep the making process steady.

🎯 Exam Tip: Distinguish process control from product control by emphasizing that it aims to "minimize defective items in the production process" and uses "control charts" to achieve this.

 

Question 7. Define a control chart.
Answer: A control chart is a drawing or graph used to show data, which helps to easily see how much a process changes from its normal standards or goals. These charts are easy to create and understand. They quickly show a manager if a process is working correctly (within limits) or if there are problems. It's a visual tool to monitor quality over time.
In simple words: A control chart is a simple graph that shows if a production process is working as it should, or if something is going wrong.

🎯 Exam Tip: Explain that control charts are "graphic devices" for "presenting data" that reveal "variations from established standards" and show if a process is "in control".

 

Question 8. Name the control charts for variables.
Answer: The control charts used for variables are:
(i) Charts for mean (\( \bar {X} \))
(ii) Charts for Range (R)
These two charts help monitor different aspects of a variable in a process.
In simple words: For things that can be measured (variables), we use charts for the average value (mean) and for how spread out the values are (range).

🎯 Exam Tip: Clearly state both \( \bar {X} \) (mean) and R (range) charts. These are fundamental control charts.

 

Question 9. Define mean chart.
Answer: The mean chart (also known as the \( \bar {X} \) chart) is used to display the average quality of samples taken from a production process. These charts help in deciding whether to accept or reject the process.
**Procedure for \( \bar {X} \) chart:**
(i) First, take samples \( X_1, X_2, X_3 \), and so on. Each sample should have "n" observations (usually \( n = 4, 5 \) or \( 6 \)).
(ii) Next, calculate the mean for each sample: \( \bar {X}_i = \frac { \Sigma X_i }{n} \), where \( i = 1, 2, 3, 4, ... \) and \( \Sigma X_i \) is the total of "n" values in that specific sample.
(iii) Finally, find the grand mean (average of all sample means): \( \bar { \bar X} = \frac { \Sigma \bar X }{ \text{number of samples} } \), where \( \Sigma \bar X \) is the total of all the sample means. This overall average helps set the center line for the chart.
In simple words: A mean chart shows the average quality of items from a production line. To make it, you take small groups of items, find their average, and then find the average of all those averages.

🎯 Exam Tip: Clearly state the purpose of the \( \bar {X} \) chart (quality average) and remember the three main steps to calculate the sample means and the grand mean for the chart.

 

Question 10. Define R chart.
Answer: The R chart is used to show the variability or spread of the samples taken from a production process. Like the mean chart, R charts are also needed for making decisions to accept or reject the process. It helps see how much the items in each sample differ from each other.
**Procedure for R-Charts:**
(i) Calculate the range \( R = X_{max} - X_{min} \) for each sample. \( X_{max} \) is the largest value and \( X_{min} \) is the smallest value in that sample.
(ii) The average range, \( \bar {R} \), is then calculated by summing all the individual ranges and dividing by the number of samples: \( \bar {R} = \frac { \Sigma R }{n} \). This average range helps to understand the typical spread of data.
In simple words: An R chart shows how much items within each group vary. You find the biggest and smallest item in each group, subtract them to get the range, then average all those ranges.

🎯 Exam Tip: For the R chart, focus on its role in indicating "variability or dispersion." The calculation \( R = X_{max} - X_{min} \) is crucial, along with finding the average range \( \bar {R} \).

 

Question 11. What are the uses of statistical quality control?
Answer: Statistical Quality Control (SQC) has several important uses:
(i) Its main job is to gather and analyze relevant data to find out if a production process is under control or not.
(ii) SQC is valuable because it can quickly spot assignable causes (specific problems) in a process. Often, these problems are found before the product even becomes faulty.
(iii) Statistical quality control is primarily a diagnostic tool. It simply tells us whether the set standard for quality is being maintained.
(iv) This method is used in almost all manufacturing industries, such as those making automobiles, textiles, electrical equipment, biscuits, soaps, chemicals, and petroleum products.
(v) SQC serves two main purposes: (a) process control and (b) product control.
The ultimate goal of SQC is to create statistical methods that help eliminate specific causes of problems and keep the production process under control.
In simple words: Statistical Quality Control helps check if a factory's process is working right, finds problems fast, and is used in many industries. It helps control both how products are made and the products themselves.

🎯 Exam Tip: When listing the uses, ensure to mention both diagnostic capability and the twin goals of process and product control. Provide a few industry examples to show broad applicability.

 

Question 12. Write the control limits for the mean chart.
Answer: The control limits for an \( \bar {X} \) chart are calculated differently based on whether the standard deviation (SD) is known or not.
**Case (i) When Standard Deviation (SD) is given:**
Upper Control Limit (UCL) \( = \bar { \bar X} + 3\frac { \sigma }{ \sqrt{n} } \)
Center Line (CL) \( = \bar { \bar X} \)
Lower Control Limit (LCL) \( = \bar { \bar X} - 3\frac { \sigma }{ \sqrt{n} } \)
**Case (ii) When Standard Deviation (SD) is not given:**
Upper Control Limit (UCL) \( = \bar { \bar X} + A_2\bar {R} \)
Center Line (CL) \( = \bar { \bar X} \)
Lower Control Limit (LCL) \( = \bar { \bar X} - A_2\bar {R} \)
Here, \( \sigma \) is the process standard deviation, \( n \) is the sample size, \( A_2 \) is a control chart constant (which depends on \( n \)), \( \bar { \bar X} \) is the grand mean, and \( \bar {R} \) is the average range. These formulas define the acceptable boundaries for the process.
In simple words: For a mean chart, the control limits are found using formulas. If you know the spread of data (standard deviation), one set of formulas is used. If you don't know it, you use a different set of formulas that include the average range.

🎯 Exam Tip: Clearly distinguish the two cases (SD given vs. SD not given) and provide both sets of formulas accurately. Understanding \( A_2 \) as a constant factor is important.

 

Question 13. Write the control limits for the R chart.
Answer: The control limits for an R chart are calculated differently based on whether the standard deviation (SD) is known or not.
**Case (i) When Standard Deviation (SD) is given:**
Upper Control Limit (UCL) \( = \bar {R} + 3\sigma_R \)
Center Line (CL) \( = \bar {R} \)
Lower Control Limit (LCL) \( = \bar {R} - 3\sigma_R \)
**Case (ii) When Standard Deviation (SD) is not given:**
Upper Control Limit (UCL) \( = D_4\bar {R} \)
Center Line (CL) \( = \bar {R} \)
Lower Control Limit (LCL) \( = D_3\bar {R} \)
Here, \( \sigma_R \) is the standard deviation of the ranges, \( \bar {R} \) is the average range, and \( D_3 \) and \( D_4 \) are control chart constants that depend on the sample size (n). These limits help determine if the process variability is under control.
In simple words: For an R chart, the control limits also change depending on if the data spread (standard deviation) is known. There are two sets of formulas using either the standard deviation of ranges or special constants \( D_3 \) and \( D_4 \) with the average range.

🎯 Exam Tip: As with the \( \bar {X} \) chart, clearly present both cases for the R chart. Ensure the correct constants (\( D_3 \), \( D_4 \)) are used when SD is not given.

 

Question 14. A machine is set to deliver packets of a given j weight. Ten samples of size five each were recorded. Below are given relevant data:

Answer: We need to calculate the control limits for both the mean chart and the range chart.
Given: Number of samples \( = 10 \), Sample size \( n = 5 \), \( A_2 = 0.58 \), \( D_3 = 0 \), and \( D_4 = 2.115 \).
First, calculate the sum of sample means \( \Sigma \bar{X} \) and sum of ranges \( \Sigma R \).
From the table:
\( \Sigma \bar{X} = 15 + 17 + 15 + 18 + 17 + 14 + 18 + 15 + 17 + 16 = 162 \)
\( \Sigma R = 7 + 7 + 4 + 9 + 8 + 7 + 12 + 4 + 11 + 5 = 74 \)
Next, calculate the grand mean \( \bar{ \bar X} \) and the average range \( \bar{R} \).
\( \bar{ \bar X} = \frac{ \Sigma \bar{X} }{ \text{no. of samples} } = \frac{162}{10} = 16.2 \)
\( \bar{R} = \frac{ \Sigma R }{ \text{no. of samples} } = \frac{74}{10} = 7.4 \)
**Control Limits for \( \bar{X} \) chart:**
UCL \( = \bar{ \bar X} + A_2\bar{R} \)
\( = 16.2 + (0.58)(7.4) \)
\( = 16.2 + 4.292 \)
\( = 20.492 \approx 20.49 \)
CL \( = \bar{ \bar X} = 16.2 \)
LCL \( = \bar{ \bar X} - A_2\bar{R} \)
\( = 16.2 - (0.58)(7.4) \)
\( = 16.2 - 4.292 \)
\( = 11.908 \approx 11.91 \)
**Control Limits for Range (R) chart:**
UCL \( = D_4\bar{R} \)
\( = (2.115)(7.4) \)
\( = 15.651 \approx 15.65 \)
CL \( = \bar{R} = 7.4 \)
LCL \( = D_3\bar{R} \)
\( = (0)(7.4) = 0 \)
So, the control limits for the mean chart are UCL \( = 20.49 \), CL \( = 16.2 \), LCL \( = 11.91 \). The control limits for the range chart are UCL \( = 15.65 \), CL \( = 7.4 \), LCL \( = 0 \). These limits help monitor the machine's performance.
In simple words: First, we find the average of all the sample means and the average of all the ranges. Then, using special numbers \( A_2, D_3, D_4 \), we calculate the upper, center, and lower limits for both the mean chart and the range chart.

🎯 Exam Tip: Clearly show all intermediate calculations for \( \Sigma \bar{X} \), \( \Sigma R \), \( \bar{ \bar X} \), and \( \bar{R} \). Double-check the multiplication with control chart constants \( A_2, D_3, D_4 \).

 

Question 15. Ten samples each of size five are difawn at regular intervals from a manufacturing process. The sample means (X) and their ranges (R) are given below:

**Calculate the control limits in respect of \( \bar {X} \) chart. (Given \( A_2 = 0.58, D_3 = 0 \) and \( D_4 = 2.115 \)) Comment on the state of control.**
Answer: We need to calculate the control limits for both the mean chart and the range chart, then comment on the process control.
Given: Number of samples \( = 10 \), Sample size \( n = 5 \), \( A_2 = 0.58 \), \( D_3 = 0 \), and \( D_4 = 2.115 \).
First, sum the sample means \( \Sigma \bar{X} \) and sum the ranges \( \Sigma R \).
From the table:
\( \Sigma \bar{X} = 49 + 45 + 48 + 53 + 39 + 47 + 46 + 39 + 51 + 45 = 462 \)
\( \Sigma R = 7 + 5 + 7 + 9 + 5 + 8 + 8 + 6 + 7 + 6 = 68 \)
Next, calculate the grand mean \( \bar{ \bar X} \) and the average range \( \bar{R} \).
\( \bar{ \bar X} = \frac{ \Sigma \bar{X} }{ \text{no. of samples} } = \frac{462}{10} = 46.2 \)
\( \bar{R} = \frac{ \Sigma R }{ \text{no. of samples} } = \frac{68}{10} = 6.8 \)
**Control Limits for \( \bar{X} \) chart:**
UCL \( = \bar{ \bar X} + A_2\bar{R} \)
\( = 46.2 + (0.58)(6.8) \)
\( = 46.2 + 3.944 \)
\( = 50.144 \approx 50.14 \)
CL \( = \bar{ \bar X} = 46.2 \)
LCL \( = \bar{ \bar X} - A_2\bar{R} \)
\( = 46.2 - (0.58)(6.8) \)
\( = 46.2 - 3.944 \)
\( = 42.256 \approx 42.26 \)
**Control Limits for Range (R) chart:**
UCL \( = D_4\bar{R} \)
\( = (2.115)(6.8) \)
\( = 14.382 \approx 14.38 \)
CL \( = \bar{R} = 6.8 \)
LCL \( = D_3\bar{R} \)
\( = (0)(6.8) = 0 \)
**Conclusion:**
For the \( \bar{X} \) chart, the limits are UCL \( = 50.14 \), CL \( = 46.2 \), LCL \( = 42.26 \).
For the R chart, the limits are UCL \( = 14.38 \), CL \( = 6.8 \), LCL \( = 0 \).
To comment on the state of control, we check if all sample means (\( \bar{X} \)) and ranges (R) fall within their respective control limits.
Looking at the given data:
All \( \bar{X} \) values (ranging from 39 to 53) are within \( 42.26 \) and \( 50.14 \), except for \( \bar{X} = 53 \) (sample 4), which is above the UCL, and \( \bar{X} = 39 \) (samples 5 and 8), which are below the LCL.
All R values (ranging from 5 to 9) are within \( 0 \) and \( 14.38 \), except for \( R = 9 \) (sample 4) which is within the limits.
Since some sample means (\( \bar{X} \)) fall outside the control limits (53 is above UCL, 39 is below LCL), the process is **not in control**. This means there are assignable causes that need to be investigated and corrected. A process must have all points within the limits to be considered in control.
In simple words: We calculated the upper, center, and lower limits for both the mean and range charts using the given data and constants. After checking, we found that some sample means were outside these limits. This means the process is not stable and has problems that need to be fixed.

🎯 Exam Tip: Remember to always include the "Comment on the state of control" part. This requires comparing each data point to the calculated control limits to determine if the process is stable or needs intervention.

 

Question 20. In a production process, eight samples of size 4 are collected and their means and ranges are given below. Construct mean chart and range chart with control limits.

Answer: For this problem, we have the following control chart factors: \( A_2 = 0.73 \), \( D_3 = 0 \), and \( D_4 = 2.282 \). First, we calculate the sum of the sample means \( (\sum \bar{X}) \) and the sum of the ranges \( (\sum R) \).
\( \sum \bar{X} = 12 + 13 + 11 + 12 + 14 + 13 + 16 + 15 = 106 \)
\( \sum R = 2 + 5 + 4 + 2 + 3 + 2 + 4 + 3 = 25 \)
The number of samples is \( k = 8 \).

Now, we find the grand mean (\( \bar{\bar{X}} \)) and the average range (\( \bar{R} \)):
\( \bar{\bar{X}} = \frac{\sum \bar{X}}{k} = \frac{106}{8} = 13.25 \)
\( \bar{R} = \frac{\sum R}{k} = \frac{25}{8} = 3.125 \) (Rounded to \( 3.12 \) for calculations as in source)

**Control Limits for the \( \bar{X} \) Chart:**
Upper Control Limit (UCL) \( = \bar{\bar{X}} + A_2 \bar{R} \)
\( = 13.25 + (0.73)(3.12) \)
\( = 13.25 + 2.2776 \)
\( = 15.5276 \approx 15.53 \)

Center Line (CL) \( = \bar{\bar{X}} \)
\( = 13.25 \)

Lower Control Limit (LCL) \( = \bar{\bar{X}} - A_2 \bar{R} \)
\( = 13.25 - (0.73)(3.12) \)
\( = 13.25 - 2.2776 \)
\( = 10.9724 \approx 10.97 \)

**Control Limits for the R Chart:**
Upper Control Limit (UCL) \( = D_4 \bar{R} \)
\( = (2.282)(3.12) \)
\( = 7.11984 \approx 7.12 \)

Center Line (CL) \( = \bar{R} \)
\( = 3.12 \)

Lower Control Limit (LCL) \( = D_3 \bar{R} \)
\( = (0)(3.12) \)
\( = 0 \)
In simple words: We calculated the upper, center, and lower limits for both the mean chart and the range chart. These limits help us see if a production process is working as it should, without too much change.

🎯 Exam Tip: Remember to use the correct \( A_2, D_3, \text{ and } D_4 \) values corresponding to the sample size \( n \) provided in the question. These values are crucial for accurate control limit calculations.

 

Question 21. In a certain bottling industry the quality control inspector recorded the weight of each of the 5 bottles selected at random during each hour of four hours in the day. Construct mean chart and range chart with control limits.

Answer: We are given the sample size \( n = 5 \) and the following control chart factors for \( n=5 \): \( A_2 = 0.58 \), \( D_3 = 0 \), and \( D_4 = 2.115 \).
First, we calculate the sum of observations \( (\sum X) \), the mean \( (\bar{X}) \), and the range \( (R) \) for each sample. Then we sum these values to find the overall totals.

Now, we calculate the grand mean (\( \bar{\bar{X}} \)) and the average range (\( \bar{R} \)):
\( \bar{\bar{X}} = \frac{\sum \bar{X}}{\text{number of samples}} = \frac{164}{4} = 41 \)
\( \bar{R} = \frac{\sum R}{\text{number of samples}} = \frac{16}{4} = 4 \)

**Control Limits for the \( \bar{X} \) Chart:**
Upper Control Limit (UCL) \( = \bar{\bar{X}} + A_2 \bar{R} \)
\( = 41 + (0.58)(4) \)
\( = 41 + 2.32 \)
\( = 43.32 \)

Center Line (CL) \( = \bar{\bar{X}} \)
\( = 41 \)

Lower Control Limit (LCL) \( = \bar{\bar{X}} - A_2 \bar{R} \)
\( = 41 - (0.58)(4) \)
\( = 41 - 2.32 \)
\( = 38.68 \)

**Control Limits for the R Chart:**
Upper Control Limit (UCL) \( = D_4 \bar{R} \)
\( = (2.115)(4) \)
\( = 8.46 \)

Center Line (CL) \( = \bar{R} \)
\( = 4 \)

Lower Control Limit (LCL) \( = D_3 \bar{R} \)
\( = (0)(4) \)
\( = 0 \)

**Conclusion:** Since all the sample mean points and range points fall within their calculated upper and lower control limits, the production process is considered to be in control. This shows the process is stable and predictable over time.
In simple words: We calculated the top, middle, and bottom lines for charts that show how much the weights of bottles change. Because all the measured weights stayed within these lines, it means the bottling machine is working smoothly and consistently.

🎯 Exam Tip: When given different data sets or sample sizes, always ensure you select the correct control chart factors (\(A_2, D_3, D_4\)) that match the given sample size \(n\) for each specific problem.