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2 years ago

Find the mean ,median and mode of the weights of people shown

Mathematics

1 answer:

Alexandra [31]2 years ago

4 0

Answer:

$(a)\ Mean = 80kg$

$(b)\ Median = 76kg$

$(c)\ Mode = 61kg$

Step-by-step explanation:

Given

$Data:105kg, 53kg ,76kg ,91kg ,120kg ,61kg ,55kg ,98kg ,61kg$

$n = 9$

Solving (a): The mean

This is calculated using:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{105kg+53kg +76kg +91kg +120kg +61kg +55kg +98kg +61kg}{9}$

$\bar x = \frac{720kg}{9}$

$\bar x = 80kg$

Hence:

$Mean = 80kg$

Solving (b): The median

$Data:105kg, 53kg ,76kg ,91kg ,120kg ,61kg ,55kg ,98kg ,61kg$

Arrange in ascending order

$Data: 53kg ,55kg, 61kg ,61kg,76kg ,91kg ,98kg , 105kg,120kg$

$n = 9$

So, the position of the median element is:

$Median = \frac{n+1}{2}th$

$Median = \frac{9+1}{2}th$

$Median = \frac{10}{2}th$

$Median = 5th$

The 5th element is 76kg

Hence:

$Median = 76kg$

Solving (c): The mode

In the given data

$Mode = 61kg$

Because it has a frequency of 2 (the highest), others have 1

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$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

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