Final answer:
(a) the mean and standard deviation of B is 19.8 ounces and 2.12 ounces.
(b) the probability that the total weight of the 4 Bartlett pears is between 18 and 22 ounces is 0.687.
(c) the mean and standard deviation of T is 93.6 ounces and 2.37 ounces.
(d) the mean and standard deviation of W is 106.6 ounces and 2.37 ounces.
Step-by-step explanation:
(a) To find the mean and standard deviation of B, we need to add together the weights of the 4 Bartlett pears.
Since the weights of Bartlett pears are normally distributed with a mean of 4.95 ounces and a standard deviation of 1.06 ounces, the mean of B is 4.95 * 4 = 19.8 ounces.
The standard deviation of B is sqrt(1.06^2 * 4)
= 2.12 ounces.
(b) To find the probability that the total weight of the 4 Bartlett pears is between 18 and 22 ounces, we need to find the z-scores corresponding to these weights.
The z-score of 18 ounces is (18 - 19.8) / 2.12
= -0.85.
The z-score of 22 ounces is (22 - 19.8) / 2.12
= 1.04.
Using a standard normal distribution table, we can find that the probability of a z-score between -0.85 and 1.04 is 0.687.
(c) To find the mean and standard deviation of T, we need to add together the weights of the 4 Bartlett pears and 8 Honeycrisp apples.
Since the weights of Bartlett pears are normally distributed with a mean of 4.95 ounces and a standard deviation of 1.06 ounces, and the weights of Honeycrisp apples are normally distributed with a mean of 8.40 ounces and a standard deviation of 1.28 ounces, the mean of T is 4.95 * 4 + 8.40 * 8
= 93.6 ounces.
The standard deviation of T is sqrt(1.06^2 * 4 + 1.28^2 * 8)
= 2.37 ounces.
(d) To find the mean and standard deviation of W, we need to add together the weights of the box, packing material, 4 Bartlett pears, and 8 Honeycrisp apples.
Since the box and packing material weigh a constant 13 ounces, the mean of W is 13 + 93.6
= 106.6 ounces.
The standard deviation of W is the same as the standard deviation of T, 2.37 ounces.