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sweet [91]
2 years ago
15

The table shows data from a survey about the number of times families eat at restaurants during a week. The families are either

from Rome, Italy or New York, New York: Maximum Minimum Q1 Q3 IQR Median Mean σ Rome 16 0 3 13 10 8.5 8 5.4 New York 20 1 4.5 6 1.5 5.5 7.25 5.4 Which of the choices below best describes how to measure the center of this data? Both centers are best described with the mean. Both centers are best described with the median. The Rome data center is best described by the mean. The New York data center is best described by the median. The Rome data center is best described by the median. The New York data center is best described by the mean.

Mathematics
1 answer:
Naily [24]2 years ago
6 0

Answer:

Both centres are best described by the median.

Step-by-step explanation:

Here is a summary of the statistics from your data.

<u>City </u>      <u>Min</u>    <u>Q1 </u>   <u>IQR</u>  <u>   Q3  </u>   <u>Max</u>     <u>Median</u>   <u>Mean</u>   <u>  σ    </u>

Rome      0     3.60  8.65  12.25     16         8.25       7.99     5.20

NY           1      2.25  4.69   6.64     20        5.45       6.39     5.91

The box plots below show that both centres are best described by the median.

The outlier in the New York data greatly distorts the mean but does not affect the median. The mean without the outlier would have been 4.45.


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It’s rational, for a number to be rational you have to be able to write it as a fraction and you can write 0.515115111511115111115... as a fraction.
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Samantha and Mia each left Julia’s house at the same time. Mia walked north at 7 kilometers per hour. Samantha ran west at 11 ki
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Mia walked 7km/h so after 1 hour, she is 7 km north of the house of Julia

Samantha walked 11km/h so after 1 hour, she is 11 km west of the house of Julia.

The points where Mia and Samantha are after 1 hour , and the house of Julia form a right triangle with sides 7 and 11 km. The distance between the girls, is the hypotenuse of his triangle.

 by the pythagorean theorem:

MS= \sqrt{ 7^{2} + 11^{2} }= \sqrt{49+121}= \sqrt{170}=  13 (km)


Answer: 13 km

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The distance between place A and place B is 60 miles. the distance from place A to place C is 15% of the distance from place A t
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A certain company's main source of income is a mobile app. The company's annual profit (in millions of dollars) as a function of
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x = $3, or x = $11


Step-by-step explanation:

The equation given is P(x)=-2(x-3)(x-11)

where

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  • x is the app price

<u>We want app prices (x's) when profit (P(x)) is 0, so plugging in into the equation:</u>

P(x)=-2(x-3)(x-11)\\0=-2(x-3)(x-11)\\0=(x-3)(x-11)

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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that
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Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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Z = \frac{46 - 40}{5.66}

Z = 1.06

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X = 34

Z = \frac{X - \mu}{\sigma}

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Z = -1.06 has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

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