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3 years ago

Negate: "In every good book there is a plot twist or surprise ending."

Mathematics

1 answer:

uysha [10]3 years ago

7 0

Answer:

"There exists a good book that does not have a plot twist and does not have a surprise ending".

Step-by-step explanation:

We negate the universal quantifier "for all" or equivalently "In every" using the existential quantifier "There exists". So, we negate "In every good book" as "There exists a good book". In the other hand, we have the propositions

P: there is a plot twist

Q: there is a surprise ending,

and the conjunction

P ∨ Q. We negate this conjunction using the De Morgan's Laws as

¬(P∨Q) = ¬P∧¬Q

i.e., does not have a plot twist and does not have a surprise ending. Therefore, we negate "In every good book there is a plot twist or surprise ending" as "There exists a good book that does not have a plot twist and does not have a surprise ending".

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$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

So the answer for this case would be n=1337 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=1.1$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =0.07 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(1.1)}{0.07})^2 =1336.006 \approx 1337$

So the answer for this case would be n=1337 rounded up to the nearest integer

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