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Luda [366]
4 years ago
12

Let X have a binomial distribution with parametersn = 25and p. Calculate each of the following probabilities using the normal ap

proximation (with the continuity correction) for the casesp = 0.5, 0.6, and 0.8and compare to the exact binomial probabilities calculated directly from the formula forb(x; n, p).(Round your answers to four decimal places.)(a)P(15 ≤ X ≤ 20)
Mathematics
1 answer:
olga55 [171]4 years ago
3 0

Answer:

The answer is explained below

Step-by-step explanation:

We have the following formulas:

from binomial distibution: P (X = x) = (nCx) * (p) x * (1-p) n-x

from normal distribution: P (X <= x) = (x-np) / sqrT (np (1-p))

Now, n = 25 and p (0.5, 0.6, 0.8), we replace in the formulas and we are left with the following table:

 P        P(15<=X<=20)                    P(14.5<=X<=20.5)

0.5            0.2117     is less than            0.2112

0.6            0.5763     is less than            0.5685

0.8            0.5738    is greater than       0.5957

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RSB [31]

Answer:

x = 5

Step-by-step explanation:

11 = 4x-9

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3 years ago
A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

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\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

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By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

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\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

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Step-by-step explanation:

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