The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives 16 years; the
standard deviation is 1.7 years.
Use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a gorilla living longer than 14.3
years.

Percent % pls

the probability of a gorilla living longer than 14.3 years is 83.9%

Given :

The lifespans of gorillas in a particular zoo are normally distributed

Mean is 16 years  and standard deviation is 1.7 years

Empirical rule diagram is attached below

We need to find the probability of a gorilla living longer than 14.3

Lets find out 14.3 lies in which standard deviation on left  or right

mean is 16

14.3 lies on first standard deviation on left of mean 16

So we find out the area that covers after 14.3

The area after 14.3 is

the probability of a gorilla living longer than 14.3 years is 83.9%

Learn more : link

Using the Emperical rule:

68% lie with one one standard deviation:

16 + 1.7 , 16-1.7 = 17.7, 14.3

14.3 is part of the 68%.

The remaining 32% of the distribution is outside the range, with half being less than and half being greater than.

32/2 = 16

The probability of living loner than 14.3 Would be 16%

10

step-by-step explanation:

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*