Answer:
Pr(X-Y ≤ 44.2) = 0.5593
Step-by-step explanation:
for a certain breed of terrier
Mean(μ) = 72cm
Standard deviation (σ) = 10cm
n = 64
For a certain breed of poodle
Mean(μ) = 28cm
Standard deviation (σ) = 5cm
n = 100
Let X be the random variable for the height of a certain breed of terrier
Let Y be the random variable for the height of a certain breed of poodle
μx - μy = 72 -28
= 44
σx - σy = √(σx^2/nx + σy^2/ny)
= √10^2/64 + 5^2/100
= √100/64 + 25/100
= √ 1.8125
= 1.346
Using normal distribution,
Z= (X-Y- μx-y) / σx-y
Z= (44.2 - 44) / 1.346
Z= 0.2/1.346
Z= 0.1486
From the Z table, Z = 0.149 = 0.0593
Φ(z) = 0 0593
The probability that the difference of the observed sample mean is at most 44.3 is Pr(Z ≤ 44.2)
Recall that if Z is positive,
Pr(Z≤a) = 0.5 + Φ(z)
Pr(Z ≤ 44.2) = 0.5 + 0.0593
= 0.5593
Therefore,
Pr(X-Y ≤ 44.2) = 0.5593
