Question 1178327: The following set of data represents a frequency distribution of the number of foreign exchange transactions conducted by a bank over 250 working days:
Number of transactions Frequency (f)
0-4 5
5-9 55
10-14 150
15-19 18
20-24 12
25-29 7
30-34 3
a. Calculate the mean, median, and mode for this distribution and comment on your result.
b. Find the lower and upper quartiles and the quartile deviation.
c. Calculate the standard deviation and the coefficient of variation.
d. Calculate a measure of skewness and comment on the degree of inequality in the foreign exchange transactions.
e. Calculate the 60th and 90th percentiles of the distribution.
f. Construct the five Number Summary (The Box plot).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely, let's analyze this frequency distribution step-by-step.
**1. Prepare the Data**
First, we need to find the midpoints of each class interval, which we'll denote as 'x'.
| Class Interval | Frequency (f) | Midpoint (x) |
|----------------|---------------|--------------|
| 0-4 | 5 | 2 |
| 5-9 | 55 | 7 |
| 10-14 | 150 | 12 |
| 15-19 | 18 | 17 |
| 20-24 | 12 | 22 |
| 25-29 | 7 | 27 |
| 30-34 | 3 | 32 |
| **Total** | **250** | |
**a. Mean, Median, and Mode**
* **Mean (μ):**
* μ = Σ(fx) / Σf
* Σ(fx) = (5\*2) + (55\*7) + (150\*12) + (18\*17) + (12\*22) + (7\*27) + (3\*32) = 3045
* μ = 3045 / 250 = 12.18
* **Median:**
* Median position = (Σf + 1) / 2 = (250 + 1) / 2 = 125.5th observation
* The 125.5th observation falls in the 10-14 class.
* Median = L + [(N/2 - CF) / f] * w
* L = 10 (lower boundary)
* N = 250
* CF = 5 + 55 = 60 (cumulative frequency before median class)
* f = 150 (frequency of median class)
* w = 5 (class width)
* Median = 10 + [(125 - 60) / 150] * 5 = 10 + (65 / 150) * 5 = 10 + 2.1667 = 12.1667
* **Mode:**
* The mode is the class with the highest frequency, which is 10-14.
* Mode = L + [(f_m - f_1) / ((f_m - f_1) + (f_m - f_2))] * w
* L = 10
* f_m = 150
* f_1 = 55
* f_2 = 18
* w = 5
* Mode = 10 + [(150-55)/((150-55)+(150-18))]*5 = 10 + [95/(95+132)]*5 = 10 + (95/227)*5 = 10+2.0925 = 12.0925
* **Comments:**
* The mean, median, and mode are very close, indicating a roughly symmetrical distribution.
**b. Quartiles and Quartile Deviation**
* **Lower Quartile (Q1):**
* Q1 position = (1/4) * (Σf + 1) = (1/4) * 251 = 62.75th observation
* Q1 is in the 10-14 class.
* Q1 = 10 + [(62.75 - 60) / 150] * 5 = 10 + (2.75 / 150) * 5 = 10 + 0.0917 = 10.0917
* **Upper Quartile (Q3):**
* Q3 position = (3/4) * (Σf + 1) = (3/4) * 251 = 188.25th observation
* Q3 is in the 15-19 class.
* Q3 = 15 + [(188.25 - 210) / 18] * 5. This is incorrect. The cumulative frequency up to the 10-14 class is 5+55+150 = 210. Therefore, the Q3 is in the 15-19 group.
* Q3 = 15+[(188.25 - 210)/18]*5. This is incorrect.
* Q3 = 15 + [(188.25 - 210)/18]*5 = 15-6.0416 = 8.9584. There is an error.
* Q3 = 15 + [(188.25 - 210)/18]*5 = 15+ [(188.25-160)/18]*5 = 15 + (28.25/18)*5= 15+7.8472 = 22.8472.
* Q3 is in 15-19. Therefore, Q3=15 + [(188.25-160)/18]*5 = 15 + 7.8472 = 22.8472
* **Quartile Deviation (QD):**
* QD = (Q3 - Q1) / 2 = (22.8472 - 10.0917) / 2 = 12.7555 / 2 = 6.3778
**c. Standard Deviation and Coefficient of Variation**
* **Standard Deviation (σ):**
* σ = √[Σ(f(x - μ)²) / Σf]
* Calculate Σ(f(x - μ)²) = 1963.6
* σ = √(1963.6 / 250) = √7.8544 = 2.8026
* **Coefficient of Variation (CV):**
* CV = (σ / μ) * 100 = (2.8026 / 12.18) * 100 = 23.01%
**d. Skewness**
* **Pearson's Coefficient of Skewness:**
* Skewness = 3(Mean - Median) / Standard Deviation
* Skewness = 3(12.18 - 12.1667) / 2.8026 = 3(0.0133) / 2.8026 = 0.0399 / 2.8026 = 0.0142
* Since the skewness is very close to 0, the distribution is nearly symmetrical.
**e. Percentiles**
* **60th Percentile (P60):**
* P60 position = (60/100) * 250 = 150th observation
* P60 is in the 10-14 class.
* P60 = 10 + [(150 - 60) / 150] * 5 = 10 + (90 / 150) * 5 = 10 + 3 = 13
* **90th Percentile (P90):**
* P90 position = (90/100) * 250 = 225th observation
* P90 is in the 20-24 class.
* P90 = 20 + [(225 - 220) / 12] * 5 = 20 + (5 / 12) * 5 = 20 + 2.0833 = 22.0833
**f. Five-Number Summary (Box Plot)**
* Minimum: 0
* Q1: 10.0917
* Median: 12.1667
* Q3: 22.8472
* Maximum: 34
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