The distribution of heights of a certain breed of terrier has a mean of 72 centimeters and a standard de- viation of 10 centimeters, whereas the distribution of heights of a certain breed of poodle has a mean of 28 centimeters with a standard deviation of 5 centimeters. assuming that the sample means can be measured to any degree of accuracy, find the probability that the sample mean for a random sample of heights of 64 ter- riers exceeds the sample mean for a random sample of heights of 100 poodles by at most 44.2 centimeters.

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

boxplot e is the answer