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Morgarella [4.7K]

3 years ago

5

C(a)=7500a−1500C, left parenthesis, a, right parenthesis, equals, 7500, a, minus, 1500, M(c) = 0.9c - 50M(c)=0.9c−50M, left pare

nthesis, c, right parenthesis, equals, 0, point, 9, c, minus, 50, y M(C(a))=6750a-1400M(C(a))=6750a−1400

1 answer:

neonofarm [45]3 years ago

8 0

Here's link to the answer:

tinyurl.com/wpazsebu

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Answer:

a) $P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)$

And we can find this probability with the complement rule:

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b) $z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}= 2.431$

And if we use the z score we got:

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Step-by-step explanation:

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

$X \sim N(106,0.3)$

Where $\mu=106$ and $\sigma=0.3$

Part a

We are interested on this probability

$P(X>106.11)$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And using this formula we got:

$P(X>106.11)=P(\frac{X-\mu}{\sigma}\frac{106.11-106}{0.3})=P(z>0.37)$

And we can find this probability with the complement rule:

$P(z>0.37)=1-P(z$

Part b

For this case we select a sample of n =44 and the new z score formula is given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score we got:

$z =\frac{106.11- 106}{\frac{0.3}{\sqrt{44}}}= 2.431$

And if we use the z score we got:

$P(z>2.431) =1-P(z$

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