The number of years a Bulldog lives is a random variable with mean 9 and standard deviation 3 , while for Chihuahuas, the mean is 15 and the standard deviation is 4 . Approximate the probability the that in a kennel of 100 Bulldogs and 100 Chihuahuas, the average Chihuahua lives at least 7 years longer than the average Bulldog.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

From the central limit theorem since the sample size for both cases are >30 we can assume that the average follows a normal distribution.

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And let's put some notation

B= represent the bulldogs, C= Chihuahuas

[tex]\bar X_B \sim N(\mu_B =9, \sigma_{\bar x_B}=\frac{3}{\sqrt{100}}=0.3)[/tex]

[tex]\bar X_C \sim N(\mu_C =15, \sigma_{\bar x_C}=\frac{4}{\sqrt{100}}=0.4)[/tex]

And the distribution for the difference of averages would be given by:

[tex]\bar X_C -\bar X_B \sim N(\mu_D = 15-9=6, \sigma_D=\sqrt{\frac{3^2}{100}+\frac{4^2}{100}}=0.5)[/tex]

And for this case we want this probability:

[tex]P(\bar X_C -\bar X_B > 7)[/tex]

And for this can use the z score given by:

[tex] z=\frac{\bar X_D - \mu_D}{\sigma_D}[/tex]

And after replace we got:

[tex]P(Z> \frac{(15-9)-7}{0.5})=P(Z>-2)[/tex]

And we can use the complement rule and we got this:

[tex]P(Z>-2) =1-P(Z<-2) = 1-0.02275= 0.97725[/tex]