We have been given that the average age of a breed of dog is 19.4 years. The distribution of their ages is normal and 20% of dogs are older than 22.8 years. We are asked to find the standard deviation.
Since 20% of dogs are older than 22.8 years, so 80% of dogs are younger than 22.8.
We will use z-score formula to solve our given problem.
[tex]z=\frac{x-\mu}\sigma}[/tex], where,
z = z-score,
x = Random sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
Now we will use normal distribution table to find the z-score corresponding to 80% or [tex]0.80[/tex].
z-score corresponding to 0.80 is 0.845.
Upon substituting our given values in z-score formula, we will get:
[tex]0.845=\frac{22.8-19.4}{\sigma}[/tex]
[tex]\sigma=\frac{22.8-19.4}{0.845}[/tex]
[tex]\sigma=\frac{3.4}{0.845}[/tex]
[tex]\sigma=4.0236[/tex]
[tex]\sigma\approx 4.02[/tex]
Therefore, the standard deviation is approximately 4.02 years.
