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MA_775_DIABLO [31]
1 year ago
15

Please help me, im struggling

Mathematics
1 answer:
Molodets [167]1 year ago
7 0

<u>Part 1</u>

<u />f(a)=4-7a+16a^2

<u>Part 2</u>

<u />f(a+h)=4-7(a+h)+16(a+h)^2\\\\=4-7(a+h)+16(a^2 +2ah+h^2)\\\\=4-7a-7h+16a^2 +32ah+16h^2

<u>Part 3</u>

<u />\frac{f(a+h)-f(a)}{h}=\frac{(4-7a-7h+16a^2 +32ah+16h^2-(4-7a+16a^2)}{h}\\\\=\frac{-7h+32ah+16h^2}{h}\\\\=32a+16h-7

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Step-by-step explanation:

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

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The standard deviation of the binomial distribution is:

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Normal probability distribution

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Z = \frac{X - \mu}{\sigma}

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 153, p = 0.62

So

\mu = E(X) = np = 153*0.62 = 94.86

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.62*0.38} = 6

Probability that greater than 96 out of 153 DVDs will work correctly.

97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{97 - 94.86}{6}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

1 - 0.6406 = 0.3594

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

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