Statistics Help Please...?

Im not sure how to do these questions so i would appreciate it if someone could do this one as an example for me.





The number of customers entering a butchers shop is recorded for a random sample of 2-minute periods with the following results:





Number of Customers 0 1 2 3 4 5 6


Number of 2-min periods 12 35 56 37 21 11 8





Find:


a) Median


b)Upper and Lower Quartiles


c)Mean


d)Sample Variance





Any help you can offer would be great. Thanx for your time!!:)





Btw 0 is meant to be above 12, 1 above 35 etc. Its a table but it wont let me do it properly here!!

Statistics Help Please...?
In this instance, the number of customers is your x and the number of 2-minute periods is the frequency with which each x occurs. For the median and your quartiles, figuring the cumulative frequency will also be helpful. The cumulative frequency is the number of x up to and including the current class being considered. For example, the cumulative frequency for x = 3 customers is 12+35+56+37,or 140 periods in which there were three or fewer customers in the store.


If x is the number of customers, f is the number of 2-minute periods c is the cumulative frequency, fx = the number of customers times the number of 2-minute periods, and fx² is the number of customers times fx, then we can make a table like this...


x. f . . c. . fx . . fx²


0 12 . 12 . . 0 . . 0


1 35 . 47. .35. .35


2 56 103 112 224


3 37 140 111 333


4 21 161 . 84 336


5 11 172 . 55 275


6 . 8 180 . 48 288





The sum of the relevant columns will be


∑f = 180 = the cumulative sum of the largest class


∑fx = 0 + 35 + 112 + 111 + 84 + 55 + 48 = 445


∑fx² = 0 + 35 + 224 + 333 + 336 + 275 + 288 = 1491





To find the median and the quartiles, you must look halfway down the list for the median and 1/4 and 3/4 for the quartiles.





If the fraction of the data points - 1/2 n for the median; 1/4 n and 3/4 n for the first and third quartiles - is not an exact integer, you take the next higher ranking number as your point. If the fraction falls on an exact integer, as it will in these cases, you average the data point and the next higher-ranking point to get the datum you want.





The raw rank of the median in this instance will be the n/2 smallest data point. This is an exact integer, namely 90, so we take the 90th smallest datum and the next larger one, or the 91st smallest, and average them. Looking at our cumulative frequencies, we see that the highest point in x = 2 is the 103rd smallest but the highest point in x = 1 is only the 47th smallest point. This means that both the 90th and 91st smallest point have the value 2. Their average is also 2, so that is the median number of people in the store.





The first quartile is 1/4 of the way through the 180 data points. This is 45 an exact integer, so we average the 45th and the next higher-ranking 46th smallest data points. In this instance, both have a value of 1, so the average is 1. This is also the value of the first quartile.





The third quartile is 3/4 of the way through the points, or at 135. This is again an exact integer, so we average the 135th and 136th smallest data point. Both of these are 3 and so is their average. The third quartile is therefore 3.





The mean is the sum of all the data points, which is just ∑fx, divided by the total number of points, which is ∑f or n. In this instance, ∑fx = 445 and ∑f = 180, so the mean is 445/180, or 2.47.





The sample variance is the same as the sample standard deviation, but without the square root sign. For data in this form, the variance s² can be most easily calculated from the formula





s² = [∑fx² − (∑fx)²/∑f] / (∑f − 1)


= (1491 − 445²/180) / (180 − 1)


= (1491 − 1100.14) / 179


= 390.86 / 179


= 2.183
Reply:median is middle number - find how many 2 min periods there are in all and find the middle one and then find the number of customers that corresponds to.


upper quartile - find out total number, find where the 3/4th number is (total * 3/4) and that number of customers is the upper quartile.


lower quartile - the same * 1/4


o get the mean , add the numbers up and divide by the number of numbers - (0 * 12) + (1 * 35) etc / total





variance is max number - min number.





Don't ask for answers, ask for methods.

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