The problem provides a dataset of average phone call durations. It requires converting this data into a grouped frequency table using Sturges' rule, plotting the less than and more than cumulative frequency curves on the same graph, calculating the geometric mean, finding the median, and determining the mode of the distribution.

1. Problem Description

The problem provides a dataset of average phone call durations. It requires converting this data into a grouped frequency table using Sturges' rule, plotting the less than and more than cumulative frequency curves on the same graph, calculating the geometric mean, finding the median, and determining the mode of the distribution.

2. Solution Steps

(i) Convert the data into a grouped frequency table using Sturges' rule.

First, we need to find the number of classes ($k$) using Sturges' rule:

$k = 1 + 3.322 * log_{10}(n)$

where $n$ is the number of observations. In this case, $n = 50$ (10 rows and 5 columns).

$k = 1 + 3.322 * log_{10}(50)$

$k = 1 + 3.322 * 1.69897$

$k = 1 + 5.6438$

$k \approx 6.6438$

Since the number of classes must be an integer, we can round $k$ to

7. So we have 7 classes.

Next, we need to find the class width ($w$):

$w = \frac{Range}{k}$

The range is the difference between the maximum and minimum values in the dataset.

Maximum value = 135

Minimum value = 12

Range = 135 - 12 = 123

$w = \frac{123}{7} \approx 17.57$

We can round the class width to

1

8.

Now, let's create the grouped frequency table:

| Class Interval | Frequency |

|----------------|-----------|

| 12 - 29 | 10 |

| 30 - 47 | 8 |

| 48 - 65 | 9 |

| 66 - 83 | 4 |

| 84 - 101 | 5 |

| 102 - 119 | 8 |

| 120 - 137 | 6 |

(ii) Plot the less than and more than cumulative frequency curves on the same graph from the grouped table.

This step would require plotting the points of the grouped frequency distribution, which cannot be done in text format.

(iii) Compute the geometric mean.

$GM = \sqrt[n]{x_1 * x_2 * ... * x_n}$

For grouped data, the geometric mean is calculated as:

$GM = \sqrt[N]{x_1^{f_1} * x_2^{f_2} * ... * x_k^{f_k}}$

Where $x_i$ is the midpoint of the class interval and $f_i$ is the frequency of that class. $N$ is the total frequency.

Class midpoints: (20.5, 38.5, 56.5, 74.5, 92.5, 110.5, 128.5)

Frequencies: (10, 8, 9, 4, 5, 8, 6)

N = 50

$log(GM) = \frac{\sum f_i * log(x_i)}{N}$

$log(GM) = \frac{10*log(20.5) + 8*log(38.5) + 9*log(56.5) + 4*log(74.5) + 5*log(92.5) + 8*log(110.5) + 6*log(128.5)}{50}$

$log(GM) = \frac{10*(1.31175) + 8*(1.58546) + 9*(1.75205) + 4*(1.87216) + 5*(1.96614) + 8*(2.04345) + 6*(2.10891)}{50}$

$log(GM) = \frac{13.1175 + 12.68368 + 15.76845 + 7.48864 + 9.8307 + 16.3476 + 12.65346}{50}$

$log(GM) = \frac{87.89}{50}$

$log(GM) = 1.7578$

$GM = 10^{1.7578}$

$GM \approx 57.25$

(iv) Compute the median.

Median $= L + (\frac{\frac{N}{2} - cf}{f}) * w$

Where $L$ is the lower boundary of the median class, $N$ is the total frequency (50), $cf$ is the cumulative frequency of the class before the median class, $f$ is the frequency of the median class, and $w$ is the class width (18).

N/2 =

2

5.

Cumulative frequencies: (10, 18, 27, 31, 36, 44, 50)

The median class is 48 - 65, since the cumulative frequency just exceeds

2

5.

$L = 47.5$ (lower boundary of the median class)

$cf = 18$ (cumulative frequency before the median class)

$f = 9$ (frequency of the median class)

$w = 18$

Median $= 47.5 + (\frac{25 - 18}{9}) * 18$

Median $= 47.5 + (\frac{7}{9}) * 18$

Median $= 47.5 + 14$

Median $= 61.5$

(v) Compute the mode of the distribution.

Mode $= L + (\frac{f_m - f_1}{2f_m - f_1 - f_2}) * w$

Where $L$ is the lower boundary of the modal class, $f_m$ is the frequency of the modal class, $f_1$ is the frequency of the class before the modal class, $f_2$ is the frequency of the class after the modal class, and $w$ is the class width.

The modal class is the class with the highest frequency.

In this case, there are several classes with the highest frequency 10, 8, 9, 4, 5, 8, 6, therefore multiple modes exist.

Since there are two classes with frequencies equal to 8, one way of approximating the mode is using the class with the overall maximum frequency. The class with the maximum frequency is the first class (12-29), therefore, we approximate that class as the mode.

$L = 11.5$

$f_m = 10$

$f_1 = 0$

$f_2 = 8$

$w = 18$

Mode $= 11.5 + (\frac{10 - 0}{2*10 - 0 - 8}) * 18$

Mode $= 11.5 + (\frac{10}{12}) * 18$

Mode $= 11.5 + 15$

Mode $= 26.5$

3. Final Answer

(i) Grouped frequency table:

| Class Interval | Frequency |

|----------------|-----------|

| 12 - 29 | 10 |

| 30 - 47 | 8 |

| 48 - 65 | 9 |

| 66 - 83 | 4 |

| 84 - 101 | 5 |

| 102 - 119 | 8 |

| 120 - 137 | 6 |

(ii) Plot the less than and more than cumulative frequency curves on the same graph from the grouped table (graphical representation needed).

(iii) Geometric Mean: approximately 57.25

(iv) Median: 61.5

(v) Mode: 26.5