Given two random variable means and standard deviations compute probability

Question

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.

Answer 1

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.

The CLT comes into play because we were not told the exact form of the distributions for lifetimes of each kind of dog. It would be a big leap to assume that the number of years an individual chihuahua lives is a normal random variable. But the CLT says if we collect together enough chihuahuas, their average lifetime will approximate a normal variable (under certain conditions that are reasonable to assume). So it is a relatively reasonable assumption that the distributions of the averages are normal.

Answer 2

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.