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#### karush

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- Jan 31, 2012

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If \(\displaystyle X \sim B(8,0.3)\), find \(\displaystyle P(X=5)\)

looks like we working off of a probability distribution table

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If \(\displaystyle X \sim B(8,0.3)\), find \(\displaystyle P(X=5)\)

looks like we working off of a probability distribution table

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This is not an area in which I am very knowledgeable, but I would say that the notation:

\(\displaystyle B(n,p)\)

implies:

\(\displaystyle P(X=k)={n \choose k}p^k(1-p)^{n-k}\)

Binomial distribution - Wikipedia, the free encyclopedia

- Jan 26, 2012

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$\displaystyle P \{X= x\} = \binom{n}{k} p^{k}\ (1-p)^{n-k}\ k=0,1,...,n\ (1)$

... and is...

$\mu = n\ p$

$\sigma= n\ p\ (1-p)\ (2)$

The problem is that if is $\mu=8$ and $\sigma = .3$ is doesn't exist any combination of n and p satisfying (2) so that may be that 'B' doesn't mean 'Binomial' in this case...

Kind regards

$\chi$ $\sigma$

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let me try this if \(\displaystyle X \sim B(8,0.3) \) and \(\displaystyle P(X=5)\)This is not an area in which I am very knowledgeable, but I would say that the notation:

\(\displaystyle B(n,p)\)

implies:

\(\displaystyle P(X=k)={n \choose k}p^k(1-p)^{n-k}\)

then plug in would be

\(\displaystyle {8 \choose 5}0.3^5(1-0.3)^{1-0.3}=0.047\)

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well, the next problem is \(\displaystyle X \sim N(15,9)\), find \(\displaystyle P(X < 16)\) so assume

$\displaystyle P \{X= x\} = \binom{n}{k} p^{k}\ (1-p)^{n-k}\ k=0,1,...,n\ (1)$

... and is...

$\mu = n\ p$

$\sigma= n\ p\ (1-p)\ (2)$

The problem is that if is $\mu=8$ and $\sigma = .3$ is doesn't exist any combination of n and p satisfying (2) so that may be that 'B' doesn't mean 'Binomial' in this case...

Kind regards

$\chi$ $\sigma$

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Yes, I get:let me try this if \(\displaystyle X \sim B(8,0.3) \) and \(\displaystyle P(X=5)\)

then plug in would be

\(\displaystyle {8 \choose 5}0.3^5(1-0.3)^{1-0.3}=0.047\)

\(\displaystyle P(X=5)={8 \choose 5}\left(\frac{3}{10} \right)^5\left(\frac{7}{10} \right)^3=\frac{583443}{12500000}\approx0.04667544\)

The normal distribution can be denoted by $N\left(\mu,\sigma^2 \right)$.well, the next problem is \(\displaystyle X \sim N(15,9)\), find \(\displaystyle P(X < 16)\) so assumeBandNare just function names.

Do you know how to standardize the data and use your table to find $P(x<16)$?

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no but assume the first step is solve for just \(\displaystyle x=16\)The normal distribution can be denoted by $N\left(\mu,\sigma^2 \right)$.

Do you know how to standardize the data and use your table to find $P(x<16)$?

thanks for all your help btw...

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- Mar 5, 2012

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Standardizing the data means calculating the so called $z$-value.no but assume the first step is solve for just \(\displaystyle x=16\)

thanks for all your help btw...

$$z=\frac{\text{boundary value of x} - \mu}{\sigma}$$

You can look up that value in a table of the

Do you have such a table?

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this is what was given:Standardizing the data means calculating the so called $z$-value.

$$z=\frac{\text{boundary value of x} - \mu}{\sigma}$$

You can look up that value in a table of thestandard normal distribution.

Do you have such a table?

If \(\displaystyle X \sim N (15,9),\) Find

a) \(\displaystyle P(X < 16)\)

b) \(\displaystyle P(X \geq 17)\)

c) \(\displaystyle P(14 \leq X \leq 15.5)\)

d) \(\displaystyle P(X > 6)\)

e) \(\displaystyle P(X>k)=0.34\), then \(\displaystyle k=?\)

so from \(\displaystyle N\left(\mu,\sigma^2 \right)\) and \(\displaystyle N (15,9)\)

\(\displaystyle z=\frac{\text{boundary value of x} - 15}{3}\)

so how do we get

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For (a) the boundary value of x is 16.If \(\displaystyle X \sim N (15,9),\) Find

a) \(\displaystyle P(X < 16)\)

so from \(\displaystyle N\left(\mu,\sigma^2 \right)\) and \(\displaystyle N (15,9)\)

\(\displaystyle z=\frac{\text{boundary value of x} - 15}{3}\)

so how do we getboundary value of x?

So \(\displaystyle z = \frac{16 - 15}{3} = \frac 1 3\).

Btw, I'm assuming your $\sigma^2=9$. There is a little ambiguity here, since it's also possible that your $\sigma=9$, which would of course give different results.

For instance from the wiki z-table, we can find that the corresponding lower tail probability is approximately 0.6293.

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from the given \(\displaystyle X-N(15,9)\) and from \(\displaystyle N(\mu,\sigma^2)\) looks like \(\displaystyle \sigma\) is \(\displaystyle 3\)Let's start with:

For (a) the boundary value of x is 16.

So \(\displaystyle z = \frac{16 - 15}{3} = \frac 1 3\).

Btw, I'm assuming your $\sigma^2=9$. There is a little ambiguity here, since it's also possible that your $\sigma=9$, which would of course give different results.

For instance from the wiki z-table, we can find that the corresponding lower tail probability is approximately 0.6293.

let me try the next one

b) \(\displaystyle P(X \geq 17)\)

so \(\displaystyle z = \frac{17 - 15}{3} = \frac {2}{3} \approx .6667\)

so from z table \(\displaystyle .7454\)

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from the given \(\displaystyle X-N(15,9)\) and from \(\displaystyle N(\mu,\sigma^2)\) looks like \(\displaystyle \sigma\) is \(\displaystyle 3\)

However, if your course is using \(\displaystyle N(\mu,\sigma)\), it becomes a different matter.

Can you tell from your course notes?

Close.let me try the next one

b) \(\displaystyle P(X \geq 17)\)

so \(\displaystyle z = \frac{17 - 15}{3} = \frac {2}{3} \approx .6667\)

so from z table \(\displaystyle .7454\)

However, in (a) we were talking about the lower tail probability.

In (b) we're talking about the upper tail probability.

The z-table you used gives the lower tail probability (see the graph next to it).

But you can calculate the upper tail probability by taking 1 minus the lower tail probability.

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OK but, I am not taking a course in this nor do I have a text or notes just trying to do a work sheet from a "summer 2014 math SL IB Questions" will be studying the subject more formerly this fall just want to get head start on it.Ifyour course is using \(\displaystyle N(\mu,\sigma^2)\), which is also on wiki, that is is correct.

However, if your course is using \(\displaystyle N(\mu,\sigma)\), it becomes a different matter.

Can you tell from your course notes?

so for b) \(\displaystyle 1-.7454 = .2546\) are you referring to the bell shaped graphClose.

However, in (a) we were talking about the lower tail probability.

In (b) we're talking about the upper tail probability.

The z-table you used gives the lower tail probability (see the graph next to it).

But you can calculate the upper tail probability by taking 1 minus the lower tail probability.

for c) we have \(\displaystyle P(14 \leq X \leq 15.5)\)

\(\displaystyle z_u = \frac{14 - 15}{3} = - \frac {1}{3} \approx -.3334\)

I presume this is upper since \(\displaystyle 14 \leq X\)

from z table \(\displaystyle -0.6293\) however this was from a neg number

and \(\displaystyle z_l=\frac{15.5 - 15}{3} = \frac {.5}{3} \approx .16667\)

from z table \(\displaystyle 0.5636\)

so \(\displaystyle -0.6293 \leq X \leq 0.5636 \)

this was a shot in the dark...

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so the last 2 questions

If \(\displaystyle X \sim N(15,9)\) find

d) \(\displaystyle P(X > 6)\)

\(\displaystyle

z_u = \frac{6 - 15}{3} = 3\)

z table \(\displaystyle = 0.9987\)

e) \(\displaystyle P(X>k)=0.34\), then \(\displaystyle k=?\)

\(\displaystyle \frac{k-15}{3}=0.34\) then \(\displaystyle k=16.02\)

If \(\displaystyle X \sim N(15,9)\) find

d) \(\displaystyle P(X > 6)\)

\(\displaystyle

z_u = \frac{6 - 15}{3} = 3\)

z table \(\displaystyle = 0.9987\)

e) \(\displaystyle P(X>k)=0.34\), then \(\displaystyle k=?\)

\(\displaystyle \frac{k-15}{3}=0.34\) then \(\displaystyle k=16.02\)

Last edited:

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e) Because 0.34 < 0.5, we know $k>\mu$. We want to find the $z$-score associated with an area of $0.5-0.34=0.16$. Once you find this $z$-score, add this many standard deviations to the mean:

\(\displaystyle k=\mu+z_k\sigma\)

What do you find?

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where does the \(\displaystyle 0.5\) come from?

e) Because 0.34 < 0.5, we know $k>\mu$. We want to find the $z$-score associated with an area of $0.5-0.34=0.16$. Once you find this $z$-score, add this many standard deviations to the mean:

\(\displaystyle k=\mu+z_k\sigma\)

What do you find?

like this?

\(\displaystyle k=15+0.8554(3)=17.5662 \)

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from thePlease refer to the following diagram:

View attachment 979

We know the red area plus the green area is $0.5$. We want the green area to be $0.34$, thus the red area must be $0.5-0.34=0.16$.

What $z$-score is associated with an area of 0.16?

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You are finding the area associated with $z=0.16$. You want to use the table the other way around.from theCumulative from mean (0 to Z)table I find \(\displaystyle 0.06356\)

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Good! And yes.so for b) \(\displaystyle 1-.7454 = .2546\) are you referring to the bell shaped graph

You're getting there...for c) we have \(\displaystyle P(14 \leq X \leq 15.5)\)

\(\displaystyle z_u = \frac{14 - 15}{3} = - \frac {1}{3} \approx -.3334\)

I presume this is upper since \(\displaystyle 14 \leq X\)

from z table \(\displaystyle -0.6293\) however this was from a neg number

and \(\displaystyle z_l=\frac{15.5 - 15}{3} = \frac {.5}{3} \approx .16667\)

from z table \(\displaystyle 0.5636\)

so \(\displaystyle -0.6293 \leq X \leq 0.5636 \)

this was a shot in the dark...

You found that $P(X < 14) \approx -0.6293$ and that $P(X \leq 15.5) \approx .16667$.

Both are lower tail probabilities and you actually want the area in between in the bell curve.

To get it, you can simply subtract them from each other, effectively removing the lower tail from the leftmost one from the lower tail of the rightmost one.

That is, $P(14 \leq X \leq 15.5) = P(X \leq 15.5) - P(X < 14) \approx .1667 - -0.6293 = 0.7960$.

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\(\displaystyle 0.15910\) is \(\displaystyle z=.41\) at least if you mean by the other way around...You are finding the area associated with $z=0.16$. You want to use the table the other way around.

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Yes, if you want a bit more accuracy, I found:\(\displaystyle 0.15910\) is \(\displaystyle z=.41\) at least if you mean by the other way around...

\(\displaystyle z_k\approx0.412463129441\)

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so then \(\displaystyle k=15+(0.412463129441)(3)=16.2374\) or round down to \(\displaystyle 16\) since has been in integersYes, if you want a bit more accuracy, I found:

\(\displaystyle z_k\approx0.412463129441\)

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