A company fills bags with fertilizer for retail sale. The weights of the bags of fertilizer have a normal distribution with a mean weight of 15 lbs and a standard deviation of 1.70 lbs. a. What is the

a. To solve this problem, we need to standardize the distribution using the z-score formula: z = (x - ?) / ?, where x is the weight of the bag, ? is the mean weight, and ? is the standard deviation. For a bag to weigh between 14 and 16 pounds, we need to find the area under the normal curve between the z-scores of (14 - 15) / 1.7 = -0.59 and (16 - 15) / 1.7 = 0.59. We can use a standard normal distribution table or a calculator to find that the probability is \approx imately 0.544. b. The distribution of sample means is also a normal distribution, with a mean of the population mean (?) and a standard deviation of the population standard deviation (?) divided by the square root of the sample size (n). In this case, the formula for the z-score is: z = (x? - ?) / (? / ?n), where x? is the sample mean weight. We know that the probability of a bag weighing between 14 and 16 pounds is 0.544. To find the probability that the average weight of 35 bags will be between 14 and 16 pounds, we need to find the z-scores corresponding to those weights. First, we need to find the mean and standard deviation of the distribution of sample means: ? = 15 ? / ?n = 1.7 / ?35 ? 0.287 Next, we standardize the distribution using the z-score formula: z = (16 - 15) / 0.287 ? 3.48 z = (14 - 15) / 0.287 ? -3.48 Using a standard normal distribution table or calculator, we can find the area between these z-scores to be \approx imately 0.0005. Therefore, the probability that the average weight of 35 bags will be between 14 and 16 pounds is \approx imately 0.0005.