Answer:
- Average age: 12.4 years old
- Standard deviation: 4.8 years
Explanation:
You want to know the mean and standard deviation of the ages of 5 zebras, given those ages are {8, 11, 17, 7, 19}.
Mean
The sum of the 5 ages is 62, so their mean is ...
average age = 62/5 = 12.4
The average age is 12.4 years old.
Standard deviation
The sum of squares of the ages is 884, so the average square is 176.8. The variance is the difference betwen this and the square of the mean:
variance = 176.8 -(12.4)² = 23.04
The standard deviation is the square root of the variance:
standard deviation = √23.04 = 4.8
The standard deviation of ages is 4.8 years.
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Additional comment
One needs to be careful when computing the standard deviation of a list of numbers. Often, variance and standard deviation formulas include a factor of n/(n-1). The purpose of this factor is to make the value be an unbiased estimate of the population standard deviation, assuming the given list is a sample from that population.
The calculation above does not include that factor, so gives the standard deviation assuming the list is the entire population. Some calculators and spreadsheets have a different function for computing this value.
In the attached, this is the difference between Sx and σx.
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