2 years ago

Please help with this question

Mathematics

1 answer:

puteri [66]2 years ago

4 0

The answer i found to be is C

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The mean per capita income is 15,451 dollars per annum with a variance of 298,116. What is the probability that the sample mean

docker41 [41]

Answer: 0.5467

Step-by-step explanation:

Let X be the random variable that represents the income (in dollars) of a randomly selected person.

Given : $\mu=15451$

$\sigma^2=298116\\\\\Rightarrow\ \sigma=\sqrt{298116}=546$

Sample size : n=350

z-score : $\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

To find the probability that the sample mean would differ from the true mean by less than 22 dollars, the interval will be

$\mu-22,\ \mu+22\\\\=15,451 -22,\ 15,451 +22\\\\=15429,\ 15473$

For x=15429

$z=\dfrac{15429-15451}{\dfrac{546}{\sqrt{350}}}\approx-0.75$

For x=15473

$z=\dfrac{15473-15451}{\dfrac{546}{\sqrt{350}}}\approx0.75$

The required probability :-

$P(15429$

Hence, the required probability is 0.5467.

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3 years ago

a ferry boat leaves the dock once per hour your waiting time for the next ferry will follow a uniform distribution from 0 to 60

olga55 [171]

Ok but what is the question

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