Descriptive Statistics in Action: A Case Study of Bongi’s Burger Scenario
Bongi, the owner of Bongi’s Burgers, is having trouble deciding how to purchase the tomato sauce and mustard for the restaurant for each month during 2024. He decides to look at his records to see how many of each item were used each month during 2023. He organises the data in the table below.
Month | Units | |
Tomato Sauce (bottles) | Mustard Sauce (bottles) | |
January | 12 | 8 |
February | 11 | 6 |
March | 6 | 4 |
April | 7 | 5 |
May | 5 | 4 |
June | 6 | 4 |
July | 7 | 21 |
August | 8 | 7 |
September | 5 | 8 |
October | 9 | 5 |
November | 9 | 6 |
December | 15 | 9 |
$$\sum_{i=1}^{n} x_i$$
$$\text{Tomato sauce} = 12 + 11 + 6 + 7 + 5 + 6 + 7 + 8 + 5 + 9 + 9 + 15 = 100 \; \text{bottles}$$
$$\text{Mustard} = 8 + 6 + 4 + 5 + 4 + 4 + 21 + 7 + 8 + 5 + 6 + 9 = 83 \; \text{bottles}$$
$$\mu = \frac{\sum_{i=1}^{n} x_i}{n}$$
$$\mu = \frac{12 + 11 + 6 + 7 + 5 + 6 + 7 + 8 + 5 + 9 + 9 + 15}{n} = 8.33$$
Yes. The supplier offers a 5% discount if Bongi orders a constant number of bottles each month. Based on the mean calculation, Bongi could order 8 bottles per month. Total for 12 months at 8 bottles/month = 8 × 12 = 96 bottles. Since this is slightly below the total usage (100), there might be a small shortfall, but with the discount, it could be worth it if the shortfall can be managed.
Based on the histogram, the data does not follow a normal distribution. The higher concentration of values in the lower ranges (with one outlier month at 15) suggests that the distribution is skewed, so it is more indicative of an exponential distribution. The data appears to be right-skewed, which is a characteristic of an exponentially distributed set of values.
Arrange the data set in ascending order:
Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Value | 5 | 5 | 6 | 6 | 7 | 7 | 8 | 9 | 9 | 11 | 12 | 15 |
Step-by-step calculation for the first quartile (Q1):
Step 1: Determine the position of (Q_1)
$$
Q_1\ \text{Position} = \frac{n + 1}{4} = \frac{12 + 1}{4} = 3.25
$$
Step 2: Find the lower and upper values surrounding the position
- Lower value at position 3: (6)
- Upper value at position 4: (6)
Step 3: Calculate the fractional leftover
$$
\text{Fractional Leftover} = 6 \; – \; 6 = 0
$$
Step 4: Find the fractional value
$$
\text{Fractional Value} = 0 \cdot (6 \; – \; 6) = 0 \cdot 0 = 0
$$
Step 5: Add the fractional value to the lower value
$$
Q_1 = 6 + 0
$$
Therefore, the first quartile (Q1) value is:
$$
Q_1 = 6.00
$$
Step-by-step calculation for the third quartile (Q3):
Step 1: Determine the position of (Q_3)
$$
Q_3\ \text{Position} = \frac {3 \cdot(n + 1)}{4} = \frac {3 \cdot(10 + 1)}{4} = 8.25
$$
Step 2: Find the lower and upper values surrounding the position
- Lower value at position 9: (9)
- Upper value at position 10: (11)
Step 3: Calculate the fractional leftover
$$
\text{Fractional Leftover} = 9.75 \; – \; 9 = 0.75
$$
Step 4: Find the fractional value
$$
\text{Fractional Value} = 0.75 \cdot (11 \; – \; 9) = 0.75 \cdot 2 = 1.5
$$
Step 5: Add the fractional value to the lower value
$$
Q_3 = 9 + 1.5 = 10.5
$$
Therefore, the third quartile (Q3) value is:
$$
Q_3 = 10.50
$$
Interquartile Range:
$$
IQR = Q_3 \; – \; Q_1 = 10.50 \; – \; 6.00 = 4.50
$$
December
- Median
- Outlier
- Coefficient
- Measure of skewness
2. Outlier
- Mean
- Median
- Mode
- Mean “Yes”
- Median “No”
- Mode