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julia-pushkina [17]
2 years ago
14

Please help me 18 points

Mathematics
2 answers:
erma4kov [3.2K]2 years ago
8 0

City B had the highest noon temperature, Interquartile Range, Larger Median temperature while City A had smaller range of temperature.

A box and whisker plot, also known as a box plot, shows a five-number summary of a set of data. The minimum, first quartile, median, third quartile, and maximum are the five-number summary. A box plot is created by drawing a box from the first to third quartiles. At the median, a vertical line runs through the box. The whiskers move from the lowest to the highest quartile.

From the figure:-

  • City B has the highest noon temperature with 86F
  • City B has the highest Interquartile Range
  • City B has the highest median noon temperature with 79F
  • City A has smaller range of temperature

Learn more about Statistics here :

brainly.com/question/23091366

#SPJ1

andrezito [222]2 years ago
6 0

Answer:

(a) B

(b) B

(c) B

(d) B

Step-by-step explanation:

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0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 20 hours, standard deviation of 6:

This means that \mu = 20, \sigma = 6

Sample of 150:

This means that n = 150, s = \frac{6}{\sqrt{150}}

What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?

This is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19.5. So

X = 21

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

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

Z = \frac{21 - 20}{\frac{6}{\sqrt{150}}}

Z = 2.04

Z = 2.04 has a p-value of 0.9793

X = 19.5

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

Z = \frac{19.5 - 20}{\frac{6}{\sqrt{150}}}

Z = -1.02

Z = -1.02 has a p-value of 0.1539

0.9793 - 0.1539 = 0.8254

0.8254 = 82.54% probability that the mean of this sample is between 19.25 hours and 21.0 hours

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