Question

HA

Jin recorded the ages (in years) for a sample of 4 zebras from a wildlife preserve. Here are the ages:
2,9, 14, 15
The mean of these ages is 7 = 10 years.
What is the standard deviation?
Round your answer to two decimal places.
8,
years​

Answer 1

5.94

Explanation:

Given the following information:

Ages of 4 zebras from a wildlife

• 2,9, 14, 15

=> the sample mean of this date is

=
`\bar X = (\sum_(i=1)^n X_i)/(n)`

=
`\bar X = (2+9+14+15)/(4)= 10`

The standard deviation can be determined by using this formula:

`s = \sqrt{(\sum_(i=1)^n (X_i-\bar X)^2)/(n-1)}`

<=>
`s = \sqrt{((2-10)^2 +(9-10)^2 +(14-10)^2 +(15-10)^2)/(4-1)} =5.94`