Question

Kevin Horn is the national sales manager for National Textbooks, Inc. He has a sales staff of 40 who visit college professors all over the United States. Each Saturday morning he requires his sales staff to send him a report. This report includes, among other things, the number of professors visited during the previous week. Listed below, ordered from smallest to largest, are the number of visits last week.
38 40 41 45 48 48 50 50 51 51 52 52 53 54 55
59 59 59 62 62 62 63 64 65 66 66 67 67 69 69

a. Determine the median number of calls.
b. Determine the first and third quartiles.
c. Determine the first decile and the ninth decile.
d. Determine the 33rd percentile.

Answer 1

Answer:

(a) median = 57

(b) Q1=50     Q3=63.5

(c)D1 = 41   D9=67

(d)P33= 51

Step-by-step explanation:

(a) The numbers are arranged in ascending order, and there are even number of calls. So the median will be the average of the middle terms.

middle terms are 55 and 59

Median = $\frac{55+59}{2}$

median= 57

(b) First quarter Q1 = $\frac{1}{4} number of calls$th

=$\frac{1}{4} *30$th

= 7.5th

so the first quarter lies in the 7.5th= (7+0.5)th term

So we multiply 0.5 by the difference of the 7th and 8th term

=0.5 *0= 0

Q1 = 50+0=50

Q3=$\frac{3}{4} number of calls$th

=$\frac{3}{4} *30$th

=22.5th = (22+0.5)th term

So we multiply 0.5 by the difference of the 22th and 23rd term

= 0.5*1=0.5

Q3= 63+0.5=63.5

(c) First decile (D1) = $\frac{1}{10} number of calls$th

=$\frac{1}{10} *30$th

= 3rd term

D1= 41

D9= $\frac{9}{10} number of calls$th

=$\frac{9}{10} *30$th

  =27th term

D9 = 67

(d) 33rd percentile =$\frac{33}{100} number of calls$th

=$\frac{33}{100} *30$th

= 9.9th term = (9 +0.9)th

Multiply 0.9 by the difference between the 9th and tenth term.

= 0.9*0 =0

P33 = 51