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?

The probability that the average weight of 2 packages randomly selected is less than 32 ounces can be calculated using the z-score formula. To find this, we first need to calculate the mean and standard deviation of the sample by taking the mean of two packages (32.4 x 2 = 64.8) and dividing it by 2 (64.8/2 = 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.

Finally, we can use a normal distribution table or calculator to find that the probability that an average weight of two packages will be less than 32 ounces is approximately 7%.