Question 1176817
Let's analyze the waiting time data for counters X and Y.

**a) Calculate Standard Deviation and Comment on Dispersion**

**1. Calculate the Midpoints**

* 0 to 5: 2.5
* 5 to 10: 7.5
* 10 to 15: 12.5
* 15 to 20: 17.5
* 20 to 25: 22.5
* 25 to 30: 27.5

**2. Calculate Mean and Standard Deviation for Counter X**

* **Mean (X̄):**
    * Σ(midpoint * frequency) / Σfrequency
    * [(2.5 * 8) + (7.5 * 10) + (12.5 * 24) + (17.5 * 19) + (22.5 * 17) + (27.5 * 12)] / (8 + 10 + 24 + 19 + 17 + 12)
    * [20 + 75 + 300 + 332.5 + 382.5 + 330] / 90
    * 1440 / 90 = 16

* **Standard Deviation (sX):**
    * sX = √[Σ(midpoint - X̄)² * frequency / (Σfrequency - 1)]
    * sX = √[((2.5-16)²*8) + ((7.5-16)²*10) + ((12.5-16)²*24) + ((17.5-16)²*19) + ((22.5-16)²*17) + ((27.5-16)²*12)] / 89
    * sX = √[(182.25 * 8) + (72.25 * 10) + (12.25 * 24) + (2.25 * 19) + (42.25 * 17) + (132.25 * 12)] / 89
    * sX = √[1458 + 722.5 + 294 + 42.75 + 718.25 + 1587] / 89
    * sX = √4822.5 / 89
    * sX = √54.18539
    * sX ≈ 7.36

**3. Calculate Mean and Standard Deviation for Counter Y**

* **Mean (Ȳ):**
    * Σ(midpoint * frequency) / Σfrequency
    * [(2.5 * 6) + (7.5 * 5) + (12.5 * 16) + (17.5 * 22) + (22.5 * 18) + (27.5 * 23)] / (6 + 5 + 16 + 22 + 18 + 23)
    * [15 + 37.5 + 200 + 385 + 405 + 632.5] / 90
    * 1675 / 90 = 18.61

* **Standard Deviation (sY):**
    * sY = √[Σ(midpoint - Ȳ)² * frequency / (Σfrequency - 1)]
    * sY = √[((2.5-18.61)²*6) + ((7.5-18.61)²*5) + ((12.5-18.61)²*16) + ((17.5-18.61)²*22) + ((22.5-18.61)²*18) + ((27.5-18.61)²*23)] / 89
    * sY = √[(260.6721 * 6) + (123.4321 * 5) + (37.3281 * 16) + (1.2321 * 22) + (15.1281 * 18) + (79.0921 * 23)] / 89
    * sY = √[1564.0326 + 617.1605 + 597.2496 + 27.1062 + 272.3058 + 1819.1183] / 89
    * sY = √4896.9740 / 89
    * sY = √55.02218
    * sY ≈ 7.42

**Comment on Dispersion:**

* Counter X: Mean ≈ 16 minutes, Standard Deviation ≈ 7.36 minutes
* Counter Y: Mean ≈ 18.61 minutes, Standard Deviation ≈ 7.42 minutes

Both counters have similar standard deviations, indicating comparable dispersion of waiting times. Counter Y has a slightly higher mean and standard deviation, suggesting slightly more variability and a longer average wait.

**b) Comment on the Shape of the Distribution (Without Coefficient of Skewness)**

* **Counter X:**
    * The highest frequency (12.5 midpoint) is in the middle of the distribution.
    * The frequencies decrease somewhat symmetrically on both sides.
    * This suggests a roughly symmetrical or slightly skewed distribution.

* **Counter Y:**
    * The highest frequencies are in the 15-20, 20-25, and 25-30 intervals.
    * This suggests a right-skewed distribution, as there are more customers with longer waiting times.

**c) Suggest Best Measures of Central Tendency and Dispersion**

* **Central Tendency:**
    * **Mean:** Appropriate for both distributions, especially if they are approximately symmetrical. However, if there is a strong skew, the median might be more representative.

* **Dispersion:**
    * **Standard Deviation:** Appropriate for both distributions, as it measures the average spread of data around the mean.
    * **Interquartile Range (IQR):** If there is a strong skew, the IQR might be a better measure of dispersion, as it is less affected by extreme values.

**In summary:**

* For both counters, the mean and standard deviation are suitable.
* If counter Y's right skew is significant, the median and IQR might be preferred.