The temperature readings from a thermocouple in a furnace fl

The temperature readings from a thermocouple in a furnace fluctuate according to a cumulative distribution function Determine the following. Round your answers to two decimal places (e.g. 98.76). (a)P(X

Solution

a)

As 803.0 C is between 800 and 810,

P(X < 803.0) = 0.1(803) - 80 = 0.3 [answer]

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b)

P(800 < X <= 803) is just the same as P(X < 803), as the values lower than 800 have probability 0 anyway.

Thus,

P(800 < X <= 803) = 0.3 [ANSWER]

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C)

P(X > 808.70) = 1 - P(X < 808.70)

= 1 - [0.1(808.70) - 80]

= 0.13 [answer]

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d)

Getting the cumulative areas,

P(X < 805.80) = 0.58
P(X < 808.70) = 0.87

Thus, the area between these two values is

P(805.80 < X < 808.70) = 0.29

hence, the area OUTSIDE these two values is

P(outside) = 1 - 0.29 = 0.71 [ANSWER]