Question

1. Use technology to compute summary statistics (mean, SD, and five number summary) for the lead levels data from problem 4.

2. Here are the data:

3. sample	lead (mg/L)
   1	0.23
   2	0.39
   3	0.17
   4	0.20
   5	0.18
   6	0.16
   7	0.25
   8	0.14
   9	0.00
   10	0.17
   11	0.00
   12	0.06
   13	0.10
   14	0.02
   15	0.09
   16	0.12

   1. What lead level is the 50th percentile?

   2. What’s the average lead level?

   3. On average, how far are the lead levels from the mean lead level?

   4. What is these lead levels’ interquartile range?

4. sample	lead (mg/L)
   1	0.23
   2	0.39
   3	0.17
   4	0.20
   5	0.18
   6	0.16
   7	0.25
   8	0.14
   9	0.00
   10	0.17
   11	0.00
   12	0.06
   13	0.10
   14	0.02
   15	0.09
   16	0.12

Answer 1

	X
	0.23
	0.39
	0.17
	0.20
	0.18
	0.16
	0.25
	0.14
	0.00
	0.17
	0.00
	0.06
	0.10
	0.02
	0.09
	0.12
MEAN	0.1425
S.D	0.0983
MIN	0
MAX	0.39
Q1	0.0675
Q2	0.15
Q3	0.195
	0.1275
	0.15

All the Mean,S.D,MIN, MAX, , , and are calculated using EXCEL.

After typing the given value in column B; from B2 to B17; with following commands we will get those values.

COMMANDS:

To get Mean; =AVERAGE(B2:B17)

To get S.D; =STDEV.P(B2:B17)

To get Minimum Value; =MIN(B2:B17)

To get Maximum Value: =MAX(B2:B17)

To get ; =QUARTILE.EXC(B2:B17,1)

To get : =QUARTILE.EXC(B2:B17,2)

To get ; =QUARTILE.EXC(B2:B17,3)

To get ; =PERCENTILE.EXC(B2:B17,0.5)

********************************************

On an Average, how far the lead levels from the mean lead level::

	X
	0.23	0.0875
	0.39	0.2475
	0.17	0.0275
	0.20	0.0575
	0.18	0.0375
	0.16	0.0175
	0.25	0.1075
	0.14	-0.0025
	0.00	-0.1425
	0.17	0.0275
	0.00	-0.1425
	0.06	-0.0825
	0.10	-0.0425
	0.02	-0.1225
	0.09	-0.0525
	0.12	-0.0225
TOTAL		0.0000
AVERAGE	0.1425

Therefore

On an Average lead levels from the mean lead level is Zero.