Were given a random variable X representing the amount of gasoline and gallons purchased by a randomly selected customer at a gas station. Yeah, we're told that the mean and standard deviation of X are 11.54 point zero. In part, they were asked to find in a sample of 50 randomly selected customers the approximate probability that the sample mean amount purchased is at least 12 gallons. So we have the X bar is the sample average amount of gas purchased by 50 customers. Now we know by the central Limit theorem, since 50 is greater than 30 supplies, we have that X bar is a approximately normal with expected value of X bar being the same as the average of the population which were given is 11.5 and the standard deviation of X bar is the standard deviation of the population. Sigma over the square it of the number of samples and and we're given that sigma is 4.0, and we have the end is 50 and so we have four over route 50 and therefore the probability that X bar is going to be at least 12. This is the same approximately as one minus Phi of 12 minus the mean 10.5 Sorry, 11.5 I mean standard deviation 4.0 over Route 50. This is approximately one minus five 0.88 And using a calculator or a table, we find that this is equal to approximately 0.1 94 0.1894 Next in Part B were given a sample of 50 randomly selected customers and rest to find the approximate probability that the total amount of gasoline purchased is at least 600 gallons. So now we're dealing with the total amount of gasoline purchased, so we'll let TBD total amount of gas purchased by 50 customers. Now we have that the customers total purchase is that least 600 gallons. Well, so T is when we greater than equal to 600. If and only if we have that customers, average purchase is going to be these at least 600 over the number of customers. 50. I'm sorry, not at least, but if their average purchase is 650 which is 12 gallons, but notice that finding the probability that T is greater for 600 this means this is finding the probability that X bar is greater than 12. So this is the same as part A. Therefore, we have that the probability that T is greater than or equal to 600 approximately equal to 0.1894 Finally, in part, C has to find the approximate value of the 95th percentile for the total amount purchased by 50 randomly selected customers. Well, as from Part B, we're keeping the variable t and we have that by the Central Limit Theorem. Since we have at least 50 customers, T is approximately normally distributed, in particular with I mean, which is number of customers 50 times the mean cost, which is 11.5. This is equal to 575 gallons in a staring deviation which is square root of the number of customers, 50 times the standard deviation of the population, which is four. This is approximately 28.28 gallons. Um, using this information, you know, the 95th percentile of the standard normal distribution is about 1.645 follows the 95th percentile of tea. This is approximately while our mean of T 575 plus 1.645 times the standard deviation, which is approximately 28.28 So 95th percentile of tea is approximately 621.5 gallons.