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3 years ago

5)

Mathematics

1 answer:

notsponge [240]3 years ago

4 0

Answer:

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Step-by-step explanation:

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Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passe

Angelina_Jolie [31]

Answer:

a) $P(X=0)=(3C0)(0.8)^0 (1-0.8)^{3-0}=0.008$

$P(X=1)=(3C1)(0.8)^1 (1-0.8)^{3-1}=0.096$

$E(X) = np = 3*0.8= 2.4$

We expect 2.4 successes in a sample of three selected.

And the standard deviation is given by:

$Sd(X)= \sigma = \sqrt{np(1-p)}=\sqrt{3*0.8*(1-0.8)}= 0.693$

Represent the typical variation around the mean.

b) $Y \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Y=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

c) $Z \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Z=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

Step-by-step explanation:

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".

Solution to the problem

Part a

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

$X \sim Binom(n=3, p=0.8)$

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:

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

And we want the following probabilities

$P(X=0)=(3C0)(0.8)^0 (1-0.8)^{3-0}=0.008$

$P(X=1)=(3C1)(0.8)^1 (1-0.8)^{3-1}=0.096$

The mean is given by:

$E(X) = np = 3*0.8= 2.4$

We expect 2.4 successes in a sample of three selected.

And the standard deviation is given by:

$Sd(X)= \sigma = \sqrt{np(1-p)}=\sqrt{3*0.8*(1-0.8)}= 0.693$

Represent the typical variation around the mean.

Part b

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

$Y \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Y=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

Part c

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

$Z \sim Binom(n=4, p=0.8)$

And we want this probability:

$P(Z=3)=(4C3)(0.8)^3 (1-0.8)^{4-3}=0.4096$

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A pony, tied to the end of a lead, can walk in a full circle, a distance of 56.55 meters. About how long is its lead?

sergejj [24]

B is the correct answer

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1.1666666667 that as all

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Probably 18 i’d say i could be wrong tho:)

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Ainat [17]

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Step-by-step explanation:

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