Question

Q,: (2 marks) An administrator wanted to study the utilization of long-distance telephone service by a department. One variable of interest (let's call it X) is the length, in minutes, of long-distance calls made during one month. There were 38 calls that resulted in a connection. The length of calls, already ordered from smallest to largest, are presented in the following table. 1.6 1.7 1.8 1.8 1.9 2.1 4.5 4.5 5.9 7.1 7.4 7.5 12.7 15.3 15.5 15.9 15. 9 16.1 19.4 22.5 23.5 24.0 31. 7 32.8 Which one of the following statements is not true? 2.5 7.7 16.5 43. 5 3.0 8.6 17.3 53.31 3.0 9.3 17. 5 4.4 9.5 19.0 A) The 75th percentile (Q) is 17.5 minutes. B) The 50th percentile is (Q.) 9.4 minutes. C) The 25th percentile (Q.) is 4.4 minutes. D) Q-Q, >Q,- Q E) Average x > Median x F) X distribution is positively skewed. G) The percentile rank of 5.9 minutes is 13. H) Range of X is 51.7 minutes.

Answer 1

Sr. No.	X
1	1.6
2	1.7
3	1.8
4	1.8
5	1.9
6	2.1
7	2.5
8	3
9	3
10	4.4
11	4.5
12	4.5
13	5.9
14	7.1
15	7.4
16	7.5
17	7.7
18	8.6
19	9.3
20	9.5
21	12.7
22	15.3
23	15.5
24	15.9
25	15.9
26	16.1
27	16.5
28	17.3
29	17.5
30	19
31	19.4
32	22.5
33	23.5
34	24
35	31.7
36	32.8
37	43.5
38	53.3
Total	508.2
Mean	13.374

A percentile means that at pth percentile, p% of the data lies below the 'pth value = X'

1. Quartile formula

$Q_{i}=\frac{(i)(1+N)}{4}th \;value$

75th percentile means 3rd Quartile

Q3 = (3*39)/4th = 29.25th = 29 + 0.25 (30 - 29)th value

Q3 = 17.875  $\neq$ 17.5

False

2.

50th percentile means2nd quartile

Q2 = 19.5th = 19th + 0.5(20-19)th value

= 9.4

True

3. 25th means Q1

Q1 = 9.75 th = 9th + 0.75 (10th - 9th)

= 4.05

  $\neq$ 4.4

False

4. Q3 - Q2 = 8.475

Q2 - Q1 = 5.35

Therefore Q3 - Q2 >Q2 - Q1

True

5.

Average = $\frac{\sum x}{n}$ = 13.374

Media = Q2 = 9.4

Average > Median

True

6.

X	Freq
0--10	20
10--20	11
20--30	3
30--40	2
40--50	1
50--60	1

0--10 1 0--20 20--30 30--40 40--50 50-60

If we look at the distribution the frequencies continues to decrease. But fora positive skewed the first class would have to have a lower freq which suddenly increases and then steeps lower gradually creating a tail towards right. Since the there is no low frequnecy in the beginning it is not positively skewed

False

6. here we want to find the rank

So 5.9 is the rank = 13th value

this is the data rank not the percentile rank

$p_{i}=\frac{i(1+N)}{100}th\;value$

13 = $\frac{i(1+N)}{100}th\;value$

i = 33

that means 5.9mins is the 33rd percentile.

False

7. Range = Max - Min

Range = 51.7

True