Answer:
[tex](a)\ Mean = 80kg[/tex]
[tex](b)\ Median = 76kg[/tex]
[tex](c)\ Mode = 61kg[/tex]
Step-by-step explanation:
Given
[tex]Data:105kg, 53kg ,76kg ,91kg ,120kg ,61kg ,55kg ,98kg ,61kg[/tex]
[tex]n = 9[/tex]
Solving (a): The mean
This is calculated using:
[tex]\bar x = \frac{\sum x}{n}[/tex]
[tex]\bar x = \frac{105kg+53kg +76kg +91kg +120kg +61kg +55kg +98kg +61kg}{9}[/tex]
[tex]\bar x = \frac{720kg}{9}[/tex]
[tex]\bar x = 80kg[/tex]
Hence:
[tex]Mean = 80kg[/tex]
Solving (b): The median
[tex]Data:105kg, 53kg ,76kg ,91kg ,120kg ,61kg ,55kg ,98kg ,61kg[/tex]
Arrange in ascending order
[tex]Data: 53kg ,55kg, 61kg ,61kg,76kg ,91kg ,98kg , 105kg,120kg[/tex]
[tex]n = 9[/tex]
So, the position of the median element is:
[tex]Median = \frac{n+1}{2}th[/tex]
[tex]Median = \frac{9+1}{2}th[/tex]
[tex]Median = \frac{10}{2}th[/tex]
[tex]Median = 5th[/tex]
The 5th element is 76kg
Hence:
[tex]Median = 76kg[/tex]
Solving (c): The mode
In the given data
[tex]Mode = 61kg[/tex]
Because it has a frequency of 2 (the highest), others have 1
