Answer:
Percentage of armadillos between 13 and 17 years = 79.052%f using Standard Normal Distribution Tables
Step-by-step explanation:
As we know from normal distribution: z(x) = (x - Mu)/SD
where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution
Therefore using given data: Mu = 14, SD = 1.2 we have z(x) by using z(x) = (x - Mu)/SD as under:
Approach 1 using Standard Normal Distribution Table:
z for x=17: z(17) = (17-14)/1.2 gives us z(17) = 2.5
z for x=13: z(13) = (13-14)/1.2 gives us z(13) = -0.83
Afterwards using Normal Distribution Tables we find the probabilities as under:
P(17) using z(17) = 2.5 gives us P(17) = 99.379%
Similarly we have:
P(13) using z(13) = -0.83 gives us P(13) = 20.327%
Finally in order to find out the probability between 17 & 13 years we have:
Percentage of armadillos between 13 and 17 years = P(17) - P(13) = 99.379% - 20.327% = 79.052%
The standard normal distribution table is being attached for yours easiness.
Approach 2 using Excel or Google Sheets:
P(17) = norm.dist(17,14,1.2,1)
P(13) = norm.dist(13,14,1.2,1)
Percentage of armadillos between 13 and 17 years = { P(17) - P(13) } * 100
( here 2y is GCF common term so taking outside the terms of the polynomial )
( now taking (-) outside )