[MUSIC] Suppose weights of checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds, which roughly corresponds to about 23 kilos. What percent of airlines passengers are expected to incur this fee? We're told that baggage weights are nearly normally distributed with mean 45 and standard deviation 3.2. So we could write out our normal model here. And to find the desired probability, one approach would be to use the applet. So let's go to the applet at this address and work through how we can actually use that to calculate the probability. We select the Distribution to be Normal. We want to set our Mean to 45, so we can slide it across to 45. And we want to set our standard deviation to 3.2. Here we can see our normal distribution being drawn. And remember that we want to find the upper tail because we're looking to see what percent of the passengers have baggages that weigh in excess of 50 pounds, so we're going to slide our cutoff value of interest to 50, and gives us that about 5.91% of passengers have, are expected to have baggages that weigh in excess of 50 pounds. Another approach is to use r directly on the p norm function that we learned earlier. So in R, we set our p-norm function. Our cutoff value is 50, mean is 45, standard deviation is 3.2, and we find that the area under the curve that's below 50 is 0.9409. Well, that is not really the area that we're interested in, so to find the desired area we need to find the complement of this answer, so 1 minus 0.9409 gets us to the same answer, 0.0591. So roughly 5.91% of the passengers are expected to have baggages that weigh in excess of 50 pounds. We can also do this calculation by hand using z-scores and the normal probability table. For that, the first thing we need to do is to calculate our Z score as the observation 50 minus the mean 45 divided by the standard deviation 3.2 which gets us to a Z score of 1.56 and then we refer to our table. Locate 1.5 on the row and 0.06 on the column. And a Z score of 1.56 then corresponds to the value at the intersection of these 0.9406. Remember, this is going to be the area under the curve below 50. And to find the area above the curve below this value, we simply need to find the complement, which once again gets us to the same value, roughly 5.9%. Another question, the average daily high temperature in June in LA is 77 degrees Fahrenheit, with a standard deviation of five degrees Fahrenheit. Suppose that temperatures in June closely follow a normal distribution. How cold are the coldest 20% of the days during June in LA? We are told that the distribution is nearly normal, with mean 77 degrees Fahrenheit, and standard deviation five degrees Fahrenheit. So we can draw our curve, and we also mark the coldest 20% of the day, so that's going at the lower end of our distribution. So the shaded area here is .20. This time we don't know our cutoff value. That's the number that we're trying to get to. So in order to do that, we need to first figure out which standardized score, or Z score this corresponds to. We can do that using r, and in for that we're going to use the other function that we learned about, the qnorm function. Where the first input is the percentile, the second input is the mean, and the third input is the standard deviation, and that gives us 72.79. Meaning that the coldest 20% of the days during June in LA, are colder than 72, 72.79 degrees Fahrenheit. Another approach would be to do this by hand. Once again, let's draw our curve. Let's mark our percentile. And then what we want to do is we want to figure out, using the table, what Z-score the 0.2 corresponds to. So we're looking in the table and we're going to look in the center to get as close to .2 as possible and it appears that the closest we can get is going to be somewhere between .1977 and .2005. Well how do we choose between these two? Let's go with the closest, so .2005. Which gives us a Z score of negative 0.84. We know that this value can also be calculated as the unknown observation, minus the mean 77 divided by the standard deviation of 5, and if we solve for X here, multiply on both sides by 5, and adding the 77, we once again get to roughly the same answer, 72.8 degrees Fahrenheit.
