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1 year ago

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A study was conducted to determine whether there were significant differences between medical students admitted through special

programs (such as retention incentive and guaranteed placement programs) and medical students admitted through the regular admissions criteria. It was found that the graduation rate was 92.5% for the medical students admitted through special programs. Round your answers to 4 decimal places. If 12 of the students from the special programs are randomly selected, find the probability that at least 11 of them graduated.

1 answer:

iris [78.8K]1 year ago

7 0

The probability that at least 11 of them graduated is 0.7739 if graduation rate is 92.5%.

Given that graduation rate is 95.2%.

We are required to find the probability that out of 12 selected candidates atleast 11 should be graduated.

Probability is basically the chance of happening an event among all the events possible. It cannot be negative. It lies between 0 and 1.

Probability=Number of items/ Total items.

Probability that atleast 11 will be graduated out of 12 selected candidates is as under:

It will be a binomial distribution.

So,

P(X>=11)= $12C_{11}(0.925)^{11} (0.075)^{1} +12C_{12}(0.925)^{12} (0.075)^{0}$

=12*0.4241*0.075+1*0.3923

=0.3816+0.3923

=0.7739

Hence the probability that at least 11 of them graduated is 0.7739 if graduation rate is 92.5%.

Learn more about probability at brainly.com/question/24756209

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5.
$f(x)=\sin x\implies f(\pi)=0$
$f'(x)=\cos x\implies f'(\pi)=-1$
$f''(x)=-\sin x\implies f''(\pi)=0$
$f'''(x)=-\cos x\implies f'''(\pi)=1$

Clearly, each even-order derivative will vanish, and the terms that remain will alternate in sign, so the Taylor series is given by

$f(x)=-(x-\pi)+\dfrac{(x-\pi)^3}{3!}-\dfrac{(x-\pi)^5}{5!}+\cdots$
$f(x)=\displaystyle\sum_{n\ge0}\frac{(-1)^{n-1}(x-\pi)^{2n+1}}{(2n+1)!}$

Your answer is off by a sign - the source of this error is the fact that you used the series expansion centered at $x=0$ , not $x=\pi$ , and so the sign on each derivative at $x=\pi$ is opposite of what it should be. I'm sure you can figure out the radius of convergence from here.

- - -

6. Note that this is already a polynomial, so the Taylor series will strongly resemble this and will consist of a finite number of terms. You can get the series by evaluating the derivatives at the given point, or you can simply rewrite the polynomial in $x$ as a polynomial in $x-2$ .

$f(x)=x^6-x^4+2\implies f(2)=50$
$f'(x)=6x^5-4x^3\implies f'(2)=160$
$f''(x)=30x^4-12x^2\implies f''(2)=432$
$f'''(x)=120x^3-24x\implies f'''(2)=912$
$f^{(4)}(x)=360x^2-24\implies f^{(4)}(2)=1416$
$f^{(5)}(x)=720x\implies f^{(5)}(2)=1440$
$f^{(6)}(x)=720\implies f^{(6)}(2)=720$
$f^{(n\ge7)}(x)=0\implies f^{(n\ge7)}(2)=0$

$\implies f(x)=50+160(x-2)+216(x-2)^2+152(x-2)^3+59(x-2)^4+12(x-2)^5+(x-2)^6$

If you expand this, you will end up with $f(x)$ again, so the Taylor series must converge everywhere.

I'll outline the second method. The idea is to find coefficients so that the right hand side below matches the original polynomial:

$x^6-x^4+2=(x-2)^6+a_5(x-2)^5+a_4(x-2)^4+a_3(x-2)^3+a_2(x-2)^2+a_1(x-2)+a_0$

You would expand the right side, match up the coefficients for the same-power terms on the left, then solve the linear system that comes out of that. You would end up with the same result as with the standard derivative method, though perhaps more work than necessary.

- - -

7. It would help to write the square root as a rational power first:

$f(x)=\sqrt x=x^{1/2}\implies f(4)=2$
$f'(x)=\dfrac{(-1)^0}{2^1}x^{-1/2}\implies f'(4)=\dfrac1{2^2}$
$f''(x)=\dfrac{(-1)^1}{2^2}x^{-3/2}\implies f''(4)=-\dfrac1{2^5}$
$f'''(x)=\dfrac{(-1)^2(1\times3)}{2^3}x^{-5/2}\implies f'''(4)=\dfrac3{2^8}$
$f^{(4)}(x)=\dfrac{(-1)^3(1\times3\times5)}{2^4}x^{-7/2}\implies f^{(4)}(4)=-\dfrac{15}{2^{11}}$
$f^{(5)}(x)=\dfrac{(-1)^4(1\times3\times5\times7)}{2^5}x^{-9/2}\implies f^{(5)}(4)=\dfrac{105}{2^{14}}$

The pattern should be fairly easy to see.

$f(x)=2+\dfrac{x-4}{2^2}-\dfrac{(x-4)^2}{2^5\times2!}+\dfrac{3(x-4)^3}{2^8\times3!}-\dfrac{15(x-4)^4}{2^{11}\times4!}+\cdots$
$f(x)=2+\displaystyle\sum_{n\ge1}\dfrac{(-1)^n(-1\times1\times3\times5\times\cdots\times(2n-3)}{2^{3n-1}n!}(x-4)^n$

By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}(-1\times\cdots\times(2n-3)\times(2n-1))(x-4)^{n+1}}{2^{3n+2}(n+1)!}}{\dfrac{(-1)^n(-1\times\cdots\tiems(2n-3))(x-4)^n}{2^{3n-1}n!}}\right|$
$\implies\displaystyle\frac{|x-4|}8\lim_{n\to\infty}\frac{2n-1}{n+1}=\frac{|x-4|}4$
$\implies |x-4|$

so that the ROC is 4.

- - -

10. Without going into much detail, you should have as your Taylor polynomial

$\sin x\approx T_4(x)=\dfrac12+\dfrac{\sqrt3}2\left(x-\dfrac\pi6\right)-\dfrac14\left(x-\dfrac\pi6\right)^2-\dfrac1{4\sqrt3}\left(x-\dfrac\pi6\right)^3+\dfrac1{48}\left(x-\dfrac\pi6\right)^4$

Taylor's inequality then asserts that the error of approximation on the interval $0\le x\le\dfrac\pi3$ is given by

$|\sin x-T_4(x)|=|R_4(x)|\le\dfrac{M\left|x-\frac\pi6\right|^5}{5!}$

where $M$ satisfies $|f^{(5)}(x)|\le M$ on the interval.

We know that $(\sin x)^{(5)}=\cos x$ is bounded between -1 and 1, so we know $M=1$ will suffice. Over the given interval, we have $\left|x-\dfrac\pi6\right|\le\dfrac\pi6$ , so the remainder will be bounded above by

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which is to say, over the interval $0\le x\le\dfrac\pi3$ , the fourth degree Taylor polynomial approximates the value of $\sin x$ near $x=\dfrac\pi6$ to within 0.000328.

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