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IgorC [24]
3 years ago
7

The phenomenon of bottle flipping has hit the nation. You design a machine that can randomly flip a bottle. You observe 2180 ran

dom, independent flips, and 325 land the correct way. Estimate the overall proportion of bottles that land correctly when flipped randomly. Use a 98% confidence level. Round everything to 3 decimal places. a) State the parameter of interest. Verify that the necessary conditions are present in order to carry out the procedure.
Mathematics
1 answer:
krek1111 [17]3 years ago
8 0

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.

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

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.

The lower limit of this interval is:

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

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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Answer:

The most tickets were written on Saturday .On Saturday 325 tickets were issued

Step-by-step explanation:

The average number of traffic tickets issued in a city on any given day  Sunday-Saturday  can be approximated by

T(x) = -6x^2 + 84x + 37

Where  x represents the number of days after Sunday

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Sunday = x=0

Monday = x=1

Tuesday = x=2

Wednesday = x=3

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Substitute  x= 0

T(x) = -6x^2 + 84x + 37\\T(x) = -6(0)^2 + 84(0) + 37\\T(x)=37

On Sunday 37 tickets were issued

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T(x) = -6x^2 + 84x + 37\\T(x) = -6(1)^2 + 84(1) + 37\\T(x)=115

On Monday 115 tickets were issued

Substitute  x= 2

T(x) = -6x^2 + 84x + 37\\T(x) = -6(2)^2 + 84(2) + 37\\T(x)=181

On Tuesday 181 tickets were issued

Substitute  x= 3

T(x) = -6x^2 + 84x + 37\\T(x) = -6(3)^2 + 84(3) + 37\\T(x)=235

On Wednesday 235 tickets were issued

Substitute  x= 4

T(x) = -6x^2 + 84x + 37\\T(x) = -6(4)^2 + 84(4) + 37\\T(x)=277

On Thursday 277 tickets were issued

Substitute  x= 5

T(x) = -6x^2 + 84x + 37\\T(x) = -6(5)^2 + 84(5) + 37\\T(x)=307

On Friday 307 tickets were issued

Substitute  x= 6

T(x) = -6x^2 + 84x + 37\\T(x) = -6(6)^2 + 84(6) + 37\\T(x)=325

On Saturday 325 tickets were issued

Hence the most tickets were written on Saturday .On Saturday 325 tickets were issued

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Given:

Required:

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