Question

Pine trees and juniper trees are common to the San Bernardino

Answer 1

Answer:

Z-score = 0.9450

Step-by-step explanation:

Step(i):-

 mean height of pine tree (x₁⁻) = 52

standard deviation of pine tree σ₁ = 3

Mean height of Jupiter tree ((x₂⁻) = 22

standard deviation of Jupiter tree  σ₂ = 5

Size of first sample n₁ = 48

size of second sample n₂ = 33

Step(ii):-

Z-score

               $Z = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{\frac{S.D^{2} }{n_{1} }+\frac{S.D^{2} }{n_{2} }} }$

              $Z = \frac{52 -22 }{\sqrt{\frac{3^{2} }{48 }+\frac{5^{2} }{33 }} }$

             Z = 0.9450