mikaylarichardson96

20.06.2020 •

Mathematics

A triangle on a coordinate plane is translated according to the rule T-3, 6(x, y) Which is another way to write this rule?

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Ответ:

karencbetancourt

23.07.2021

Mathematics

the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4% is 0.0042

Step-by-step explanation:

Given that :

A film distribution manager calculates that 9% of the films released are flops

Let p be the probability for the movies that were released are flops;

$\mu_p = P = 0.9$

If the manager is right, what is the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4%

now; we know that our sample size = 442

the standard deviation of the variance is $\sigma_p= \sqrt{\dfrac{p(1-p)}{n}}$

$\sigma_p= \sqrt{\dfrac{0.9(1-0.9)}{442}}$

$\sigma_p= \sqrt{\dfrac{0.9(0.1)}{442}}$

$\sigma_p= \sqrt{\dfrac{0.09}{442}}$

$\sigma_p= \sqrt{2.0361991 \times 10^{-4}}$

$\sigma _p = 0.014$

So; if the manager is right; the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4% can be calculated as:

P(|p-P|0.04)=1 -P(p-P|

$P(|p-P|0.04)=1 -P(-0.04 \leq p-P \leq 0.04)$

$P(|p-P|0.04)=1 -P( \dfrac{-0.04}{\sigma_p} \leq \dfrac{ p-P}{\sigma_p} \leq \dfrac{0.04}{\sigma_p})$

$P(|p-P|0.04)=1 -P( \dfrac{-0.04}{0.014} \leq Z\leq \dfrac{0.04}{0.014})$

$P(|p-P|0.04)=1 -P( -2.8571 \leq Z\leq 2.8571)$

$P(|p-P|0.04)=1 -[P(Z \leq 2.8571) -P (Z\leq -2.8571)$

P(|p-P|0.04)=1 -(0.9979 -0.0021)

P(|p-P|0.04)=1 -0.9958

$\mathbf{P(|p-P|0.04)=0.0042}$

∴

the probability that the proportion of flops in a sample of 442 released films would differ from the population proportion by greater than 4% is 0.0042

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