ECO-07 Solved Question Paper

Question 1 – Meaning and Use of Statistics in Business

Q1. What do you understand by “Statistics”? How can statistics help in business and management decisions?

Answer

Meaning of statistics (concept):
Statistics is a way of collecting, organising, summarising and analysing numerical data so that we can make sense of it and take better decisions. Instead of relying on guesses or isolated examples, statistics helps us look at patterns in numbers.

Why statistics matters in business and management:

1. Better decisions, less guesswork
   A shopkeeper doesn’t order stock blindly. By looking at past monthly sales, they estimate how many units to order next month. That is a simple use of averages and trends.
2. Understanding customers
   Companies use surveys and data analysis to find out what price people are willing to pay, which features they like, and which age group buys more. This is all statistical work.
3. Planning production and inventory
   Factories use statistics to forecast demand, plan how much to produce, and control stock levels so that they neither over-produce nor lose sales due to shortages.
4. Quality control
   In manufacturing, a small sample of items from each batch is tested. If the defect rate is too high, the process is corrected. Statistical quality control charts are used for this.
5. Risk and uncertainty
   Finance departments use probability and past data to judge risks—such as chances of default on loans or likely movement in sales.
6. Performance measurement
   Managers track KPIs (Key Performance Indicators) like average sales per employee, customer satisfaction scores etc. These are all statistical summaries that support performance reviews.

From my own experience doing a college project on a canteen, simple stats like average spending per student and busiest time slots helped the canteen manager plan staff shifts and menu items more sensibly.

Question 2 – Errors in Sampling and the Idea of Dispersion

Q2(a). Differentiate between sampling errors and non-sampling errors in surveys.

Answer

When we collect data from only a part of the population (a sample), the results can differ from the true value of the whole population. The differences can come from two broad kinds of errors:

1. Sampling errors

These arise because we observe only a sample, not everyone.

• Even if we select the sample carefully, the sample average may not be exactly equal to the population average.
• For example, if a college has 2000 students and we survey only 100 about pocket money, their average may be a little higher or lower than the true college average.

Sampling error can be reduced by:

• increasing sample size
• using better sampling methods (random sampling, stratified sampling etc.).

2. Non-sampling errors

These are errors not due to the size of the sample, but due to other issues such as:

• Faulty questionnaire: confusing questions, leading questions
• Response bias: respondents hiding or exaggerating answers
• Recording/processing mistakes: wrong entries, coding errors, data entry mistakes
• Non-response: some people in the sample do not respond at all

These errors can affect both sample surveys and complete censuses. A badly designed question paper can ruin the data even if we asked everyone in the population.

In my own survey for a class project, we saw non-sampling error when some students simply ticked the same option all through the form to finish quickly. Their responses had to be cleaned.

Q2(b). What is meant by “dispersion” in statistics?

Answer

Dispersion tells us how spread out the data are around the centre (such as mean or median). Two sets of marks can have the same average but very different spreads.

For example, suppose two batches of students have the same average marks: 60.

• Batch A: most students score between 55 and 65.
• Batch B: some score 20, some 95, and only a few around 60.

Here, Batch B has higher dispersion.
Measures of dispersion include range, quartile deviation, mean deviation and standard deviation. Knowing dispersion helps managers understand risk and stability—steady sales month after month are less risky than sales that jump wildly up and down.

Question 3 – Short Notes

Q3. Write brief notes on the following:

(a) Advantages of using personal interviews for collecting data
(b) Limitations of using secondary data
(c) Why we classify raw data into groups

Answer

(a) Advantages of personal interviews

1. Flexibility: The interviewer can explain questions, repeat or clarify when the respondent is confused.
2. Higher response rate: People are more likely to answer when spoken to directly, compared to postal questionnaires.
3. Better quality of answers: The interviewer can ask follow-up questions and probe deeper.
4. Use with less-educated respondents: Even if respondents cannot read well, they can still answer.
5. Observation: The interviewer can observe body language and surroundings, which sometimes give extra information.

I used personal interviews to collect data from shopkeepers in a local market; most of them would not have replied to an online form.

(b) Limitations of secondary data

Secondary data are data that were collected by someone else for another purpose (e.g., government reports, company records). Problems include:

1. Different purpose: The data may not exactly suit our present study.
2. Unknown reliability: We may not know how carefully the data were collected.
3. Outdated information: Old data may not reflect current conditions.
4. Lack of detail: The data might be in broad totals, whereas we need finer breakdowns.

So secondary data are useful as a starting point, but they usually need checking and sometimes updating.

(c) Why classify data into groups

Classification means organising raw data into categories or classes (e.g., income groups, age groups). It is done to:

1. Make data manageable: Long lists of numbers become easier to see and understand.
2. Show patterns: Grouped data reveal distributions, trends and frequencies.
3. Prepare for analysis: Calculation of averages, graphs, dispersion etc. is easier after classification.
4. Better presentation: Tables and diagrams based on classified data are more readable in reports.

When I first looked at 200 individual salary figures in Excel, it felt chaotic. After grouping them into salary ranges (0–10k, 10–20k etc.), patterns became clear immediately.

Question 4 – Histogram and Frequency Polygon

Q4. The following table shows the distribution of a variable. Explain clearly how you would draw a histogram and, using the same diagram, a frequency polygon. (Use the data as given.)

| Class interval | Frequency |
| 10–20 | 12 |
| 20–30 | 30 |
| 30–40 | 35 |
| 40–50 | 65 |
| 50–60 | 45 |
| 60–70 | 25 |
| 70–80 | 18 |

Answer

Step 1: Prepare class boundaries
Since all classes are of equal width (10 units) and continuous, you can take them as:
10–20, 20–30, 30–40, 40–50, 50–60, 60–70, 70–80

Step 2: Draw the histogram

1. On the horizontal axis (X-axis), mark the class intervals (10–20, 20–30, …, 70–80).
2. On the vertical axis (Y-axis), mark a suitable scale for frequencies (e.g., 1 cm = 10 units of frequency).
3. For each class interval, draw a rectangle whose base is the class interval and height equals its frequency.
   • 10–20: rectangle up to 12
   • 20–30: up to 30
   • 30–40: up to 35
   • 40–50: up to 65
   • 50–60: up to 45
   • 60–70: up to 25
   • 70–80: up to 18
4. Make sure the rectangles are touching each other (continuous classes).

This gives the histogram.

Step 3: Draw the frequency polygon on the same graph

1. Find the midpoint of each class:
   • 10–20 → 15
   • 20–30 → 25
   • 30–40 → 35
   • 40–50 → 45
   • 50–60 → 55
   • 60–70 → 65
   • 70–80 → 75
2. Mark a point at each class midpoint at a height equal to its frequency (e.g., point at (15,12), (25,30), …, (75,18)).
3. Join these points by straight lines in order.
4. To complete the polygon, join the first point down to the mid-point before 10 (which is 5) on the X-axis at frequency 0, and the last point down to mid-point after 80 (which is 85) at frequency 0.

In practice, when I tried this on graph paper for the first time, seeing the tallest bar at 40–50 (frequency 65) and the polygon peaking there made the “shape” of the data much clearer than just looking at the table.

Question 5 – Median of a Grouped Distribution

Q5. For the frequency distribution below, compute the median value.

| Class interval | Frequency |
| 0–10 | 10 |
| 10–20 | 15 |
| 20–30 | 30 |
| 30–40 | 25 |
| 40–50 | 25 |
| 50–60 | 20 |

Answer

Step 1: Find total frequency (N)

So

Step 2: Cumulative frequencies

| Class | Frequency | Cumulative frequency |
| 0–10 | 10 | 10 |
| 10–20 | 15 | 25 |
| 20–30 | 30 | 55 |
| 30–40 | 25 | 80 |
| 40–50 | 25 | 105 |
| 50–60 | 20 | 125 |

We look for the class where the cumulative frequency first becomes greater than or equal to 62.5.

• Up to 20–30: 55
• Up to 30–40: 80

So the median class is 30–40.

Step 3: Use the median formula for grouped data

Here
L=30L = 30 (lower limit of median class)
cf=55 (cumulative frequency before median class)
f=25 (frequency of median class)
h=10 (class width)

So:

Therefore the median of the distribution is 33.

Once I understood this formula, I could get quick summaries of long grouped tables in assignments, instead of trying to locate the middle value by hand.

Question 6 – Geometric Mean and Harmonic Mean

Q6. Explain what is meant by the Geometric Mean and the Harmonic Mean. Give simple numerical examples for each and mention where they are useful.

Answer

Geometric Mean (GM)

If we have n positive values x1,x2,…,xn the geometric mean is:

It is useful for growth rates that multiply over time.

Example: An investment grows with yearly factors 1.10, 1.20 and 0.95 over three years. The geometric mean growth factor is:

This tells us the “average” multiplicative growth per year. In financial planning, I have found GM gives a more realistic long-term growth picture than a simple arithmetic average of yearly percentages.

Harmonic Mean (HM)

For n positive values x1,x2,…,xn the harmonic mean is:

HM is especially useful when dealing with rates, such as speed or cost per unit.

Example: A car travels the same distance at 40 km/h going and 60 km/h coming back. The average speed is not the simple average (40+60)/2. Instead:

So the correct average speed is 48 km/h. This example helped me realise that whenever the same distance is covered at different speeds, HM is the right tool.

Question 7 – Corrected Mean and Standard Deviation

Q7. The mean of 10 observations is 17 and the variance is 33. Later we realise that one observation, 26, was wrong and must be removed. Find the new mean and the new standard deviation of the remaining observations.

Answer

Let the original number of observations be n=10.

Total of all observations:

Using the variance relation:

So:

Now remove the wrong observation 26.

New number of observations:

New sum:

New sum of squares:

New variance:

New standard deviation:

So the corrected mean is 16, and the corrected standard deviation is approximately 5.16.

This kind of “reverse calculation” is handy in exams and in practice, when you only know the mean and variance but need to adjust for an error.

Question 8 – Karl Pearson’s Coefficient of Skewness

Q8. From the following grouped data, calculate Karl Pearson’s coefficient of skewness.

| Class interval | Frequency |
| 50–60 | 15 |
| 60–70 | 18 |
| 70–80 | 17 |
| 80–90 | 30 |
| 90–100 | 40 |
| 100–110 | 20 |
| 110–120 | 10 |

Answer

Total frequency:

Step 1: Mean

Find midpoints mmm: 55, 65, 75, 85, 95, 105, 115.
Compute ∑(f×m) :

Step 2: Mode

The highest frequency is 40 in class 90–100, so 90–100 is the modal class.

Let
L=90 (lower limit of modal class)
f1=40 (frequency of modal class)
f0=30 (frequency of previous class, 80–90)
f2=20 (frequency of next class, 100–110)
h=10 (class width)

Mode formula for grouped data:

Substituting:

Step 3: Standard Deviation

Use:

Step 4: Karl Pearson’s Coefficient of Skewness

So Karl Pearson’s coefficient of skewness is approximately –0.44, indicating a moderately negatively skewed distribution (the tail is towards the lower values).

When I plotted this distribution, I noticed the heavier frequencies on the higher-value side and a longer tail towards the lower end, which matches the negative skewness result.