Answer to Question #308582 in Statistics and Probability for Dilshaad

Question

The following data give the numbers of orders received for a sample of 30 items at the Time-saver Mail Order Company.

34 44 31 52 41 47 38 35 32 39

28 24 46 41 49 53 57 33 27 37

30 27 45 38 34 46 36 30 47 50

Using 6 classes of equal width, construct a grouped frequency distribution of the above data. Let 22 be the

lower limit of the initial class.

Question 2 (5 Marks)

Calculate the median.

Question 3 (6 Marks)

Calculate the mid-70% range.

Question 4 (11 Marks)

Calculate the coefficient of variation and interpret the value obtained.

Answer 1

Question 1

Question 2

Median $=L+((N⁄2-pcf)/f)i$

Median class is $N/2=30/2=15th$ observation, thus median class is $34-40$

Median $=34+((15-9)/8)6=34+4.5=38.5$ orders

Question 3

mid-70% = 85th percentile - 15th percentile

85th percentile = $L+((85n⁄100-pcf)/f)i=46+((25.5-21)/6)6=46+4.5=50.5$

15th percentile $=28+((4.5-3)/6)6=28+1.5=29.5$

mid-70% range $=50.5-29.5=21$ orders

Question 4

Mean $=1188/30=39.6$

standard deviation $\sigma=\sqrt{(2425.2/30)}=\sqrt{80.84}=8.9911$

coefficient of variation $=8.9911/39.6×100=22.71$ %

this means the orders are more consistent or more uniform.