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

Question

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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Eloise 4 weeks 2021-09-19T16:51:17+00:00 1 Answer 0

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    2021-09-19T16:52:59+00:00

    Answer:

    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 <76) = P(X \leq 75)

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

    “=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 <76)

    And using the continuity correction we have this:

    P(X <76)= P(X<76.5)

    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 <76) = P(X \leq 75)

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

    “=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 <76)

    And using the continuity correction we have this:

    P(X <76)= P(X<76.5)

    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

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