Find the mean, median, mode, population standard deviation...

Find the mean, median, mode, population standard deviation and variance of the continuous frequency distribution: Class Intervals 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 2 3 10 1 4 Mean = 26 Median = 24.5 Mode = 23.9 Standard deviation = 11.8 Variance = 139 Mean = 26 Median = 25 Mode = 23.9 Standard deviation = 12 Variance = 144 Mean = 26 Median = 24.5 Mode = 23.9 Standard deviation = 13 Variance = 169 Mean = 26 Median = 24.5 Mode = 23.9 Standard deviation = 14 Variance = 196

Transcript

00:01 So for this problem, the first thing that i'm going to do is convert the data from this class and frequency type table into a table where we have the midpoint and then the probability of that particular outcome.

00:15 So the first we would have a midpoint of five and also i'm going to note that i need to calculate out what the total number of data or a total number of data points is.

00:24 So two plus three is five plus ten that's 15 plus five.

00:28 So it's 20.

00:29 So we'd have 2 over 20 for a midpoint of 5, or that's 1 and 10, so 0 .1.

00:35 Then midpoint of 15 would be 3 over 20 equals 0 .15.

00:43 25, that's 10 over 20 or 0 .5.

00:46 32, 39 becomes 35.

00:49 So that has a probability of 0 .05.

00:53 And then for 45, that's 4 over 20, which is equal to 0 .2.

00:58 So now that we have those, we can find the mean value, the mean value is going to be equal to the sum of the product of the probability of each midpoint value times that midpoint value.

01:14 So we'd have 5 times 0 .1 plus 0 .15 times 15 plus 0 .5 times 25 plus 0 .05 times 25, plus 0 .05 times 45, which gives us a mean value of 26.

01:30 Then for the median that's going to be the data point such that there are that would be we want nine to the left nine less actually let's see here i'll say that we want the number less than the median would have to be equal to nine the number more than the median would have to be equal to nine so we'll have two different values theoretically which we would then take the average of.

02:05 So we have one, two, three, four, five.

02:09 Then we have that the median would have to be somewhere in this 20 to 29 range.

02:16 So actually, a more efficient way of doing this would be that the median is going to be value such that, or it's equal to the lower value of the median interval plus an total number of observations over two minus the cumulative frequency of the preceding classes c over the frequency of the particular class times h the class size.

02:46 So plugging in our values, we can see that, as i said, the median class is going to be 20.

02:52 So this is going to be equal to 20 plus n over 2 is going to be 10 minus c...

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