Question

**ANSWER BOTH QUESTIONS

1. The following observations for the respirable particles were obtained in houses located in the Midwest:
   

House #	Observed Mean Concentration (μg/m3)	Predicted Concentration (μg/m3)
1	9.9	19.1
2	4.2	6.1
3	3.9	6.2
4	3.3	5.9
5	10.2	48.8
6	7.9	40.7
7	9.4	31.2
8	11.2	26.1
9	11.8	33.9
10	11.5	15.7
11	46.0	68.4
12	48.0	66.2
13	32.0	40.2
14	26.1	30.3
15	23.1	23.9
16	37.3	47.7

The right hand column of the above table includes information on predicted concentrations using an IAQ model. Compute the correlation coefficient (r) between observed and predicted values.

1. 0.20
2. 0.30
3. 0.80
4. 0.90

2. Estimate the amount of moisture added to your house in one day period due to the following:

Four occupants

Three showers (Each occupant take bath once in a day)

One breakfast, lunch, and dinner cooked on a gas stove

Hand dishwashing after breakfast, lunch, and dinner

10 house plants

One load of laundry with the electric drier not vented outdoors.

Category	Moisture Source	Amount of Moisture released(kg)
Bathing	Shower / Occupant  / day	0.25
Washing clothes	-	0
Drying clothes	Vented outdoors / day	0
	Not vented outdoors/ day	2.565
Cooking	Breakfast (avg. of 1 person)	0.0425
	Lunch (avg. of 1 person)	0.0625
	Dinner (avg. of 1 person)	0.145
Dishwashing by hand	Breakfast (avg. of 1person)	0.025
	Lunch (avg. of 1 person)	0.02
	Dinner (avg. of 1 person)	0.08
Plant	One house plant /day	0.425
Respiration	(avg. of 1 person)in 24 hr	1.26

1. 10.3 kg
2. 14.4 kg
3. 16.4 kg
4. 20.6 kg

Answer 1

QUESTION 1: house X Y മി 4 6.2 ന് ന് : 7.9 9.4 11.2 11.8 11.5 46 48 32 26.1 23.1 37.3 XLYX^ 2y^2 19.1 189.09 98.01 364.81 6.1 25.62 17.64 37.21 24.18 15.21 38.44 5.9 19.47 10.89 34.81 48.8 497.76 104.04 2381.44 40.7 321.53 62.41 1656.49 31.2 293.28 88.36 973.44 26.1 292.32 125.44 681.21 33.9 400.02 139.24 1149.21 15.7 180.55 132.25 246.49 68.4 3146.4 2116 4678.56 66.2 3177.6 2304 4382.44 40.2 1286.4 1024 1616.04 30.3 790.83 681.21 918.09 23.9 552.09 533.61 571.21 47.7 1779.21 1391.29 2275.29 14 sum X Sum Y Sum (X.Y) Sum X^2 Sum Y^2 295.8 510.4 12976.35 8843.6 22005.18 (Sum X)^2 87497.64 (Sum Y)^2 260508.2 And we know that correlation cofficient (r) is calculated as --- r= __ [n[XX.Y) - (SX.SY V[{nX - (EX)} {nXY - (Y)"}] On putting the values and solving we get -- r= 0.8055 So Option C is right answer QUESTION 2: So according to given data there are 4 people in the house- Calculating amount of moisture accoring to it: Amount of moisture= 4*shower + 4*(Cooking breakfast+lunch+dinner) + 4*respiration 4*(dishwash breakfast+lunch+dinner) + 10*plant+ laundary+ Amount of moisture = 4*0.25 + 4*(0.0425+0.0625+0.145) + 4*1.26 + 4*(0.025+0.08+0.02) + 10*0.425 +2.565 Amount of moisture = 14.355 kg So, Option B is correct..