Fire load (MJ/m2) is the heat energy that could be released per square meter of floor area by combustion of contents and the structure itself. The following cumulative percentages are for fire loads in a sample of 388 rooms:
Fire load (MJ/m2) is the heat energy that could be
Value 0 150 300 450 600 750 900
Cumulative % 0 19.0 37.0 61.9 76.5 86.3 93.0
Value 1050 1200 1350 1500 1650 1800 1950
Cumulative % 95.0 97.8 98.1 99.5 99.6 99.8 100.0
(a) What proportion of fire loads are less than 600? At least 1200? (Round your answers to three decimal places.)
(b) What proportion of the loads are between 600 and 1200? (Round your answer to three decimal places.)

Question

11-02-2020
Mathematics

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Fire load (MJ/m2) is the heat energy that could be released per square meter of floor area by combustion of contents and the structure itself. The following cumulative percentages are for fire loads in a sample of 388 rooms:
Fire load (MJ/m2) is the heat energy that could be
Value 0 150 300 450 600 750 900
Cumulative % 0 19.0 37.0 61.9 76.5 86.3 93.0
Value 1050 1200 1350 1500 1650 1800 1950
Cumulative % 95.0 97.8 98.1 99.5 99.6 99.8 100.0
(a) What proportion of fire loads are less than 600? At least 1200? (Round your answers to three decimal places.)
(b) What proportion of the loads are between 600 and 1200? (Round your answer to three decimal places.)

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khansawaqar7 khansawaqar7 · Answer 1 · 2020-02-12T14:27:56+01:00

Answer:

a)

The proportion of fire loads are less than 600 is 0.765

The proportion of fire loads at least 1200 is 0.022

b)

The proportion of the loads between 600 and 1200 is 0.213

Step-by-step explanation:

The cumulative percent is given for various values of x.

Value (x) Cumulative percent

0 0

150 0.19

300 0.37

450 0.619

600 0.765

750 0.863

900 0.93

1050 0.95

1200 0.978

1350 0.981

1500 0.995

1650 0.996

1800 0.998

1950 1

a)

The proportion of fire loads are less than 600

P(x<600)=F(600)=0.765

Because we are given the cumulative percentages and the cumulative percentage for 450 will represent the proportion of fire loads that are less than 600.

The proportion of fire loads at least 1200

P(X≥1200)=1-F(1200)=1-0.978=0.022

b)

The proportion of the loads between 600 and 1200

P(600≤X≤1200)=?

P(a≤X≤b)=F(b)-F(a)

P(600≤X≤1200)=F(1200)-F(600)=0.978-0.765=0.213