00:01
Dear student, here is the simple random sample of 50 m &m plain candies.
00:09
We have to determine the shape of the distribution of the weight of m &m by drawing frequency histogram.
00:18
And we have to find the mean, which measure of the central tendency is better described in the weight of the plain.
00:28
So, at the beginning let's find the mean first.
00:35
Is mean that is equal to summation x divided by number of sample.
00:46
So as we can see in this sample, we have to just for example 0 .87 plus 0 .91 plus dot dot all the data plus the end is 0 .87 and this equals to 43 .73.
01:16
So we will divide this with the number is 50 and we get 43 divided by 50.
01:27
Then mean is equal to 43 .73 divided by 50 that is equal to 0 .8746.
01:42
So the mean is 0 .8746.
01:46
To find the median, first we have to check the sample size.
01:57
The sample size is 50.
01:59
50 mean even.
02:00
If the sample size is even, then we can divide 50 by 2 that is 25.
02:14
It mean 25th and 26th plus 1 we also number will be the median.
02:25
So for that we have to arrange the number in ascending order.
02:30
For example, 0 .79 is the smallest number then 0 .81 then 0 .82 repeated number is also written then go to 0 .94 and 0 .95.
02:51
In between these numbers there on 25th and 26th will be the median.
02:57
That is 0 .87 and 0 .88.
03:08
To find the median, we have to find the mean of these two numbers.
03:16
0 .87 will add with 0 .88 and divided by 2 and we get that 0 .87.
03:27
So the median is equal to 0 .87.
03:38
Now from the given above data, we have to make a frequency distribution.
03:44
Then it will be easy to draw the histogram.
03:48
So the frequency distribution to find the frequency, we have to count the number in between 0 .79 to 0 .82.
03:59
There is just one number and from 0 .82 to 0 .85 there is 10 number.
04:08
0 .85 to 0 .88 is 13.
04:12
0 .88 to 0 .91 is 14.
04:17
0 .91 to 0 .94 is 9...