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geniusboy [140]
3 years ago
13

And, S2

Mathematics
1 answer:
dangina [55]3 years ago
4 0

<u>Correct Question</u>

In the table shown, the sum of each row is shown to the right of the row and the sum of each column is shown below the column. What is the value of L?

\left|\begin{array}{c|c|c|c}-&-&-&-\\J&K&J&5\\K&K&L&13\\L&J&L&15\\-&-&-&-\\11&7&15\end{array}\right

Answer:

L=7

Step-by-step explanation:

From the first row: 2J+K=5

Therefore: K=5-2J

From the second column, 2K+J=7

Substitute K derived above into 2K+J=7

2K+J=7

2(5-2J)+J=7

10-4J+J=7

-3J=7-10

-3J=-3

J=1

Recall: K=5-2J

K=5-2(1)=3

K=3

From the third column, J+2L=15

1+2L=15

2L=15-1=14

L=7

Therefore, the value of L=7

CHECK:

\left|\begin{array}{c|c|c|c}-&-&-&-\\1&3&1&5\\3&3&7&13\\7&1&7&15\\-&-&-&-\\11&7&15\end{array}\right

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Answer:

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

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

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

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

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And we got 0.1633

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

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

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We want this probability:

P(X

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

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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 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

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

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