00:01
So we're looking at economy versus first class, and we know the mean and the standard deviation, and we have normal distributions for each of these, so we'll be able to use z values.
00:11
And we want to find in the first question, what's the likelihood that a random passenger has a, from economy, excuse me, has a mean weight between 32 and 36 pounds for the luggage.
00:25
So we want to convert those to z values, the number minus the mean divided by the standard deviations, and when i did that, i found negative 0 .8 to negative 0 .4 standard deviations below the mean and found the area below each of those z values and subtracted to find the area between.
00:44
Now, in the next question, we want to deal with first class passengers and we want to know what's the likelihood that the mean of a first class passenger from a sample size of 36, that that weight is higher than a certain amount is 0 .025.
01:01
So we want that upper tail and find out what this cutoff point is for a mean that comes from a sample size of 16.
01:09
And so this z value would correspond with a 1 .96.
01:14
And so the number minus the mean divided by that standard deviation, that standard air, will should equal this z value.
01:23
And doing the algebra, i find out that that value is 32 .94 pounds.
01:29
That, again, 2 .5 % of means of 16 first -class passengers would be this higher than this amount.
01:39
Now, on the next one, this one is a little bit more confusing.
01:43
They said that they believe they claim that it's higher than 60 % for people having luggage that is more than 35 pounds for those first -class passengers.
01:54
And we want to know what is the likelihood if we sample this of getting this type of of thing or more extreme.
02:01
That's the way i take my take on this question.
02:03
And so the sample proportion was 18 out of 36 were this weight, which only 50%.
02:10
So they're claiming it was at least 60%.
02:14
In fact, they said higher than, and i'm going to assume that it's equal to 60 % of passengers have more than 35 pounds weight.
02:23
And so the likelihood of getting what we got or more extreme, which is in the tail, converting it to a z value, comes out to be a z value of negative 1 .22.
02:33
So if the proportion of a sampling distribution is truly 60%, the likelihood of getting 50 % or more extreme, something lower than that.
02:45
So they have maybe about 11 % chance of being right...