The ages of armadillos are normally distributed, with a mean of 14 years and a standard deviation of 1.2. approximately what percentage of the armadillos are between 13 and 17 years old?

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

circle area = pi * radius^2

if diameter = 10 then radius = 5

circle area = pi * 5^2

circle area = pi * 25

circle area = 78.5398163397 square units

step-by-step explanation:

slope= 1

y-inter slope= 0

x       y

o     0

1       1

2     2

ect.

step-by-step explanation: