Question

The temperature readings from a thermocouple in a furnace fluctuate according to a cumulative distribution function,
0, x < 800 oC
F(x) = 0.05x - 40, 800 oC< or is equal to x < 820 oC
1, x > 820 oC
Determine the following.
A) P(X< 810.0).
B) P(800 C) P(X> 808.40).
D) If the specifications for the process require that the furnace temperature be between 805.60 and 808.40oC, what is the probability that the furnace will operate outside of the specifications?

Answer 1

Answer:

$P(x<810)=0.5$

$P(x<800) = 0$

$P(x>808.40) = 0.58$

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

Step-by-step explanation:

Given:

$F(x) = \left\{\begin{array}{ll}{0} &; {x&lt;800^{\circ} \mathrm{C}} \\ {0.05 x-40} &; {800^{\circ} \mathrm{C} \leq x&lt;820^{\circ} \mathrm{C}} \\ {1} &; {x&gt;820^{\circ} \mathrm{C}}\end{array}$

Solving (a): P(x < 810)

To solve this, we make use of:

$P(x<810) = F(x) = 0.05x - 40$

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:

$P(x<810)=0.5$

Solving (b): P(x < 800)

To solve this, we make use of:

$P(x<800) = F(x) = x$

Because:

$x < 800^\circ C$

So:

$P(x<800) = 0$

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:

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

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