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3 years ago

9

A family is going on a trip for 3 days. The total cost for the hotel is $369. One hundred dollars a day was budgeted for entert

ainment. How much will each day of the trip cost?

1 answer:

saul85 [17]3 years ago

6 0

Answer:

223

Step-by-step explanation:

368 divided by 3= 123

123+100= 223

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Answer:

The variable of interest is the proportion of flips that land the correct way when flipped randomly.

The necessary conditions $n\pi \geq 10$ and $n(1-\pi) \geq 10$ are present.

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Step-by-step explanation:

Variable of Interest:

Proportion of flips that land the correct way when flipped randomly.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Necessary conditions:

The necessary conditions are:

$n\pi \geq 10$

$n(1-\pi) \geq 10$

You observe 2180 random, independent flips, and 325 land the correct way.

This means that $n = 2180, \pi = \frac{325}{2180} = 0.149$

Necessary conditions

$n\pi = 2180*0.149 = 325 \geq 10$

$n(1-\pi) = 2180*0.851 = 1855 \geq 10$

The necessary conditions $n\pi \geq 10$ and $n(1-\pi) \geq 10$ are present.

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.33$ .

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$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.149 - 2.33\sqrt{\frac{0.149*0.851}{2180}} = 0.131$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.149 + 2.33\sqrt{\frac{0.149*0.851}{2180}} = 0.167$

The 98% confidence interval for the overall proportion of bottles that land correctly when flipped randomly is (0.131, 0.167).

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Step-by-step explanation:

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Step-by-step explanation:

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