A z-score of 23.9 years must be located in the gorillas' lifespans in a specific zoo's typical normal distribution.
The equation can be used to determine the z score.
z = (x - μ)/σ
In which,
Then,
z = (23.9 - 20.8)/3.1
z = 1
According to the empirical rule, 68% of lifespans are within one standard deviation of the mean.
Half of it, 68/2 = 34% %, is on the right side of the mean's standard deviation.
Given that the likelihood of a gorilla lasting less than 23.9 years is 50%.
A gorilla's lifespan being less than the norm means is;
50% + 34% = 84% sits below z-score 1.
Thus, the probability of a gorilla living less than 23.9 years is 84%.
To know more about the normal distribution, here
https://brainly.com/question/23418254
#SPJ4
The correct question is-
The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives 20.8 years; the
standard deviation is 3.1 years.
Use the empirical rule (68 – 95 - 99.7%) to estimate the probability of a gorilla living less than 23.9 years.