Suppose the weights of the contents of cans of mixed nuts have a normal distribution with mean 32.4 ounces and standard deviation 0.4 ounce.  If two packages are randomly selected, what is the probability that the average weight is less than 32 ounces?

Calculate using the Z-score Formula the probability that two randomly selected packages have an average weight below 32 ounces. For this we need to first calculate the standard deviation and mean by taking the average of two packages (32,4 x 2, which equals 64.8). Then divide it by 2. (64.8/2, or 32.4). The standard deviation would be 0.4 / √2 which equals 0.28.

We then use this information to calculate the z-score as follows: z = (x – μ)/σ where x is 32, μ is 32.4 and σ is 0.28, giving us a z-score of -1.43.

Finaly, using a normal-distribution table or calculator we can find out that the likelihood that an average package weight will be below 32 ounces for two is about 7%.