# Given two random variable means and standard deviations compute probability

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.

Initially I drew a normal curve with the chihuahua distribution with the mean 15 and proceeded to find the probability that the age exceed 16 but that is giving me a very small number of .006 which does not sound correct to me. The hint is to use central limit theory. Thanks for the help.

There is 100 chihuahuas and 100 bulldogs, the sample mean of the bulldogs follows a normal distributions with mean 9 and standard deviation $$3/\sqrt{100} = 0.3$$ and the sample mean of the age of the chihuahuas is 15 with standard deviation $$4/10 = 0.4$$. So the difference between the ages of the chihuahua and the age of the bulldogs is a normal distribution (the sum of two normal variables is normal) with mean $$15 - 9 = 6$$ and stamdard deviation $$\sqrt{\frac{16 + 9}{100}} = 0.5$$.
So in the end you have to calculat $$P(X > 7)$$ where $$X \sim \mathcal{N}(6, 0.5)$$ of $$P((X - 6)/0.5 > 2)$$ and since $$(X - 6)/0.5$$ has a normal distributions with unit variance and zero mean you can get that from your probability tables which is $$0.0228$$ so still pretty small but you should not forget that you are looking at a quite large sample here.
The chihuahuas don’t have to average $$16$$ years. The bulldogs could average $$8$$ years or less. Or the bulldogs could average $$\leq8.5$$ and the chihuahuas $$\geq15.5$$ It is even possible (though less likely) that the bulldogs average less than $$7$$ while the chihuahuas average at least $$14.$$
If $$X$$ is the average life of the $$100$$ chihuahuas and $$Y$$ the average life of the $$100$$ bulldogs, you want the probability that $$X-Y\geq 7.$$ I would not expect this to be a high probability.