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The table below shows the weights (in kg) of members in a sports club. Calculate the following:

1. The mean, median, and mode of the distribution.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Masses (kg) & [tex]$40-49$[/tex] & [tex]$50-59$[/tex] & [tex]$60-69$[/tex] & [tex]$70-79$[/tex] & [tex]$80-89$[/tex] & [tex]$90-99$[/tex] \\
\hline
Frequency & 6 & 8 & 12 & 14 & 7 & 3 \\
\hline
\end{tabular}

2. Calculate the semi-interquartile range.

3. Determine the variance.

4. Determine the standard deviation.

5. Determine the 70th percentile.

6. Determine the [tex]$t^{\text{th}}$[/tex] decile.



Answer :

GinnyAnswer
Certainly! Let's go through each of the parts of the question step by step:

### Given Information

We have the following data for masses and their corresponding frequencies:

| Masses (kg) | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
|-------------|-------|-------|-------|-------|-------|-------|
| Frequencies | 6 | 8 | 12 | 14 | 7 | 3 |

We will use the midpoints of each mass range to perform our calculations.

### Midpoints of Ranges
| Mass Ranges | Midpoints |
|-------------|-----------|
| 40-49 | 45 |
| 50-59 | 55 |
| 60-69 | 65 |
| 70-79 | 75 |
| 80-89 | 85 |
| 90-99 | 95 |

Frequencies: [6, 8, 12, 14, 7, 3]
Midpoints: [45, 55, 65, 75, 85, 95]

### 1. Calculations

#### (i) Mean, Median, and Mode

1. Mean (Average):
[tex]\[ \text{Mean} = \frac{\sum (f \times m)}{\sum f} \][/tex]
Using the given frequencies and midpoints:
[tex]\[ \sum (f \times m) = (6 \times 45) + (8 \times 55) + (12 \times 65) + (14 \times 75) + (7 \times 85) + (3 \times 95) = 410 + 440 + 780 + 1050 + 595 + 285 = 3560 \][/tex]
[tex]\[ \sum f = 6 + 8 + 12 + 14 + 7 + 3 = 50 \][/tex]
Therefore:
[tex]\[ \text{Mean} = \frac{3560}{50} = 71.2 \][/tex]

2. Median:
The median is the value separating the higher half from the lower half. To find it, we look for the middle value in the cumulative frequency distribution:
[tex]\[ \text{Cumulative frequencies: } [6, 14, 26, 40, 47, 50] \][/tex]
The total frequency is 50, so the median is the 25th and 26th values.
From the table, we see that the median class is the 60-69 range, corresponding to 65 kg.

3. Mode:
The mode is the value that appears most frequently. The highest frequency is 14, which corresponds to the 70-79 range, mid-point 75 kg.

So, the mean is [tex]\(68.4\)[/tex], the median is [tex]\(65\)[/tex], and the mode is [tex]\(75\)[/tex].

#### (ii) Semi-Interquartile Range (SIR)

1. Quartiles:
[tex]\[ Q1 \text{ (25th percentile) corresponding to cumulative frequency index of } 12.5 \][/tex]
From the cumulative frequency distribution, [tex]\(Q1\)[/tex] is found to be in the class 60-69 with midpoint [tex]\(65\)[/tex].

[tex]\[ Q3 \text{ (75th percentile) corresponding to cumulative frequency index of } 37.5 \][/tex]
From the cumulative frequency distribution, [tex]\(Q3\)[/tex] is found to be in the class 80-89 with midpoint [tex]\(85\)[/tex].

2. SIR:
[tex]\[ \text{SIR} = \frac{Q3 - Q1}{2} = \frac{85 - 65}{2} = 10 \][/tex]

#### (iii) Variance

1. Variance ([tex]\(\sigma^2\)[/tex]):
[tex]\[ \sigma^2 = \frac{\sum f(m - \text{mean})^2}{\sum f} \][/tex]
Using the mean of [tex]\(68.4\)[/tex]:

#### (iv) Standard Deviation (SD)
[tex]\[ \sigma = \sqrt{\sigma^2} = 13.8 \][/tex]

#### (v) Percentile
1. 70th Percentile (P70):
[tex]\[ P70 \text{ corresponds to cumulative frequency index of } 35. \][/tex]
The value lies in the class 70-79 with midpoint 75 kg.

#### (vi) Decile
1. 10th Decile (D10):
[tex]\[ D10 \text{ corresponds to cumulative frequency index of } 5. \][/tex]
The value lies in the class 40-49 with midpoint 45 kg.

### Summary of Results
1. Mean: [tex]\(71.2\)[/tex]
2. Median: [tex]\(65\)[/tex]
3. Mode: [tex]\(75\)[/tex]
4. Semi-Interquartile Range: [tex]\(10\)[/tex]
5. Variance: [tex]\(190.44\)[/tex]
6. Standard Deviation: [tex]\(13.8\)[/tex]
7. 70th Percentile (P70): [tex]\(75\)[/tex]
8. 10th Decile (D10): [tex]\(45\)[/tex]

These values summarize the statistical analysis for the given data distribution.

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