Mathematics, 12.03.2020 02:12 NeonPlaySword

You have a coin that is not weighted evenly and therefore is not a fair coin. Assume the true probability of getting heads when the coin is flipped is 0.52 Find the probability that less than 76 out of 157 flips of the coin are heads.

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Answer from: kiki1718

$X \sim Binom(n=157, p=0.52)$

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 this probability:

P(X

And we can use the following Excel code to find the exact

"=BINOM.DIST(75,157,0.52,TRUE)"

And we got 0.1633

The other way to solve the problem is using the normal approximation

We need to check the conditions in order to use the normal approximation.

$np=157*0.52=81.64 \geq 10$

$n(1-p)=157*(1-0.52)=75.36 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=157*0.52=81.64

$\sigma=\sqrt{np(1-p)}=\sqrt{157*0.52(1-0.52)}=6.26$

We want this probability:

P(X

And using the continuity correction we have this:

P(X

We can use the z score given by this formula $Z=\frac{x-\mu}{\sigma}$ .

$P(X< 76.5)=P(\frac{X-\mu}{\sigma}< \frac{76.5-81.64}{6.26})=P(Z < -0.821)=0.206$

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

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

$X \sim Binom(n=157, p=0.52)$

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 this probability:

P(X

And we can use the following Excel code to find the exact

"=BINOM.DIST(75,157,0.52,TRUE)"

And we got 0.1633

The other way to solve the problem is using the normal approximation

We need to check the conditions in order to use the normal approximation.

$np=157*0.52=81.64 \geq 10$

$n(1-p)=157*(1-0.52)=75.36 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=157*0.52=81.64

$\sigma=\sqrt{np(1-p)}=\sqrt{157*0.52(1-0.52)}=6.26$

We want this probability:

P(X

And using the continuity correction we have this:

P(X

We can use the z score given by this formula $Z=\frac{x-\mu}{\sigma}$ .

$P(X< 76.5)=P(\frac{X-\mu}{\sigma}< \frac{76.5-81.64}{6.26})=P(Z < -0.821)=0.206$

Answer from: Quest

step-by-step explanation:

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Answer from: Quest

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step-by-step explanation:

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