You have found the following ages (in years) of all 5 zebras at your local zoo: 8, 11, 17, 7, 19? What is the standard deviation? Round your answers to the nearest tenth.

Respuesta :

Answer:

[tex]\bar X = \frac{8+11+17+7+19}{5}= 12.4[/tex]

And the can be founded with this formula:

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}[/tex]

[tex] s = \sqrt{\frac{(8-12.4)^2 +(11-12.4)^2 +(17-12.4)^2 +(7-12.4)^2 +(19-12.4)^2}{5-1}} =5.367[/tex]

And rounded to the nearest tenth we got [tex] s= 5.4[/tex]

Step-by-step explanation:

For this case we have the following data given:

8, 11, 17, 7, 19

The first step is calculate the sample mean with this formula:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

[tex]\bar X = \frac{8+11+17+7+19}{5}= 12.4[/tex]

And the deviation can be founded with this formula:

[tex] s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}[/tex]

And replacing we got:

[tex] s = \sqrt{\frac{(8-12.4)^2 +(11-12.4)^2 +(17-12.4)^2 +(7-12.4)^2 +(19-12.4)^2}{5-1}} =5.367[/tex]

And rounded to the nearest tenth we got [tex] s= 5.4[/tex]

Answer:

Average Age 12.4

Variance 23

Step-by-step explanation:

I got it wrong and this was the answer on Khan Academy