Answer:
5.94
Step-by-step explanation:
Given the following information:
Ages of 4 zebras from a wildlife
=> the sample mean of this date is
= [tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
=[tex]\bar X = \frac{2+9+14+15}{4}= 10[/tex]
The standard deviation can be determined by using this formula:
[tex]s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}[/tex]
<=> [tex]s = \sqrt{\frac{(2-10)^2 +(9-10)^2 +(14-10)^2 +(15-10)^2}{4-1}} =5.94[/tex]
