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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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1a) 199

1b) 193

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2c) 9.5

2d) ↑

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You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

In this case the ordered data would be: 4,6,8,8,9,11,12,14,14,16.

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4) ↑

5)

You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.

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2 years ago
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Answer:

a) P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

b) P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

c) P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=10, p=0.32)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

Part b

P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]

P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211  

P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668

Part c

P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)

P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211

P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264

P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218

P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123

P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816

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