You have found the following ages (in years) of all 666 lions at your local zoo: 13,2,1,5,2,7 What is the average age of the lions at your zoo? What is the standard deviation? Average age: _____ years old Standard deviation: ____ years

Respuesta :

Answer:

[tex]Mean = 5[/tex]

[tex]S_x = 4.123[/tex]

Step-by-step explanation:

Given

Number of Lions, n: 6

Ages: 13, 2, 1, 5, 2, 7

Required

Determine the:

1. Mean

2. Standard Deviation

Mean is calculated as;

[tex]Mean = \frac{\sum x}{n}[/tex]

[tex]Mean = \frac{13+2+1+5+2+7}{6}[/tex]

[tex]Mean = \frac{30}{6}[/tex]

[tex]Mean = 5[/tex]

Standard Deviation is calculated as follows

[tex]S_x = \sqrt{\frac{\sum (x_i - Mx)^2}{N}}[/tex]

Where Mx represent mean

Substitute values for x, Mean and Land

[tex]S_x = \sqrt{\frac{(13 - 5)^2+(2 - 5)^2+(1 - 5)^2+(5 - 5)^2+(2 - 5)^2+(7 - 5)^2}{6}}[/tex]

[tex]S_x = \sqrt{\frac{(8)^2+(- 3)^2+(-4)^2+(0)^2+(-3)^2+(2)^2}{6}}[/tex]

[tex]S_x = \sqrt\frac{64+9+16+0+9+4}{6}}[/tex]

[tex]S_x = \sqrt\frac{102}{6}}[/tex]

[tex]S_x = \sqrt{17}[/tex]

[tex]S_x = 4.123[/tex]

The mean and standard deviation is 5 and 4.123 respectively

We want to find the mean or average and the standard deviation of the given set.

The average age is 5 years old and the standard deviation is 4.52 years old.

We know that for a general set of N elements {x₁, x₂, ..., xₙ} the average or mean is given by:

[tex]M = \frac{x_1 + x_2 + ... + x_n}{N}[/tex]

And the standard deviation is given by:

[tex]S = \sqrt{\frac{(x_1 - M)^2 + ... + (x_n - M)^2}{N - 1}[/tex]

The given set is:

{13, 2, 1, 5, 2, 7}

Now we just need to use the two given formulas for our set.

The mean is:

[tex]M = \frac{13 + 2 + 1+ 5 + 2 +7}{6} = 5[/tex]

And the standard deviation is:

[tex]S = \sqrt{\frac{(13 - 5)^2 + (2 - 5)^2 + (1 - 5)^2 + (5 - 5)^2 + (2 - 5)^2 + (7 - 5)^2}{6 - 1} } = 4.52[/tex]

So the average age is 5 years old and the standard deviation is 4.52 years old.

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