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lesya692 [45]
3 years ago
9

The temperature readings from a thermocouple in a furnace fluctuate according to a cumulative distribution function,

Mathematics
1 answer:
Oliga [24]3 years ago
5 0

Answer:

P(x

P(x

P(x>808.40) = 0.58

P(x\le805.40\ or\ x\ge 808.40) = 0.85

Step-by-step explanation:

Given:

F(x) = \left\{\begin{array}{ll}{0} &; {x

Solving (a): P(x < 810)

To solve this, we make use of:

P(x

Because

800^\circ C \le 810 \le 820^\circ C

So:

F(x) = 0.05 * 810 -40

F(x) = 40.5 -40

F(x) = 0.5

Hence:

P(x

Solving (b): P(x < 800)

To solve this, we make use of:

P(x

Because:

x < 800^\circ C

So:

P(x

Solving (c): P(x > 808.40)

Here, we make use of complement rule:

P(x>808.40) = 1 - P(x \le 808.40)

Where:

P(x\le808.40) = 0.05x - 40

P(x\le808.40) = 0.05*808.40 - 40

P(x\le808.40) = 0.42

So:

P(x>808.40) = 1 - 0.42

P(x>808.40) = 0.58

Solving (d) The probability that the temperature lies outside 805.60 and 808.40

First, we calculate the probability that it lies within.

This is represented as:

P(805.40 < x < 808.40)

This is calculated using:

P(805.40 < x < 808.40) = F(808.40) - F(805.40)

P(805.40 < x < 808.40) = 0.05*808.40-40 - (0.05*805.40-40)

P(805.40 < x < 808.40) = 0.42 - (0.27)

P(805.40 < x < 808.40) = 0.42 - 0.27

P(805.40 < x < 808.40) = 0.15

Using complement rule, the required probability is:

P(x\le805.40\ or\ x\ge 808.40) = 1 - 0.15

P(x\le805.40\ or\ x\ge 808.40) = 0.85

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