The guayule plant, which grows in the southwestern United States and in Mexico, is one of several plants that can be used as a source of rubber. In a large group of guayule plants, the heights of the plants are normally distributed with a mean of 12 inches and a standard deviation of 2 inches. a. What percent of the plants are taller than 16 inches? $\%$ b. What percent of the plants are at most 13 inches? $\%$ c. What percent of the plants are between 7 inches and 14 inches? $\%$ d. What percent of the plants are at least 3 inches taller than or at least 3 inches shorter than the mean height? $\%$

The correct answer to the question for (a)  the percentage of the plants which are taller than 16 inches is 2.28 % (b)the percent of the plants are at most 13 inches is 69.15% (c)the percent of the plants that are between 7 inches and 14 inches is 83.51% (d) the percent of the plants that are at least 3 inches taller than or at least 3 inches shorter than the mean height is 1

Step-by-step explanation:

Solution

Recall that

(a)μ = 12 and б = 2

Then,

Z = x- μ/б

= 16- 12 /2 = 2

thus,

1-p (z<2) = 1- 0.9772 = 0.0228

= 2.28 %

Therefore the percentage of the plants which are taller than 16 inches is 2.28 %

(b)  Z =  13-12/2

which is = 0.5

p (z<0.5)  = 0.6915

=69.15%

Therefore the percent of the plants are at most 13 inches is 69.15%

(c) we calculate for the percent of the plants that are between 7 inches and 14 inches

Now,

p(7<x<14) = p (7- 12/2 <z< 14-12/2)

p = (-2.5<z<1)

which is, =0.8351

=83.51%

(d)What percent of the plants are at least 3 inches taller than or at least 3 inches shorter than the mean height

Now,

Z = x- μ/б = 3-12/2 = -4.5

Thus,

p(z>-4.5) = p(z>-4.5)

= 1

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.