Question

The following is a stem-and leaf display for revenue from a small neighborhood Delicatessen. 50|0 9 51|0 8 9 52|0 0 3 3 9 9 53|1 2 2 3 6 54|2 3 55|1 3 Find the first and third quartiles ands the median for this data. Without calculating, guess at the mean of this data and explain your guess.

Answer 1

Data values by using stem and leaf i.e 50|0 = 500

500 , 509 , 510 , 518 , 519 , 520 , 520 , 523 , 523 , 529 , 529 , 531 , 532 , 532 , 533 , 536 , 542 , 543 , 551 , 553

Step :1 Find the median ( Q2)

The median is the middle most value of the data set

So, we get middle value of the data set is average of 10th and 11th data value , hence we get median ( Q2) = (529 +529) /2 = 529

Answer : Median = 529

Find the first quartile ( Q1)

The first quartile, denoted by Q₁ , is the median of the lower half of the data set. This means that about 25% of the numbers in the data set lie below Q₁ and about 75% lie above Q₁ .

Hence we get, average of 5th and 6th data values, which is

(519 + 520 ) / 2 = 519.5

Answer : first quartile ( Q1) = 519.5

Find the third quartile ( Q3)

The third quartile, denoted by Q₃ , is the median of the upper half of the data set. This means that about 75% of the numbers in the data set lie below Q₃ and about 25% lie above Q₃

Hence we get , average of 15th and 16th data value, which is

(533 + 536 ) / 2 = 534.5

Answer : third quartile (Q3) = 534.5

Find the mean :

Mean is the sum of all data values / number of data values

Mean = (500 + 509 + 510 + .... + 543 + 551 + 553) / 20 = 527.65

Without calculating, the mean for this data is near by median. So is approximately median

Answers :

First quartile : 519.5

Third quartile : 534.5

Median : 529

Mean : 527.65