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vodka [1.7K]
3 years ago
15

The following lists show the free throw percentages of 16 players for the Detroit Pistons and Miami Heat during the 2012-2013 NB

A season.
Detroit Pistons: 0.69, 0.73, 0.80, 0.89, 0.78, 0.81, 0.81, 0.37, 0.77, 0.62, 0.55, 0.84, 0.75, 0.83, 0.30, 0.72

Miami Heat: 0.75, 0.72, 0.81, 0.76, 0.90, 0.75, 0.82, 0.67, 0.72, 0.72, 0.57, 1.00, 0.56, 0.50, 0.50, 1.00

Calculate the median and interquartile range for each set of data. Make sure to label your calculations.
Compare the data sets using your calculations from part 1. Write your answer in complete sentences.
Mathematics
2 answers:
Tanya [424]3 years ago
8 0

Answer:

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05

Step-by-step explanation:

irina [24]3 years ago
4 0

Answer:

<u><em>Sorry if this is late</em></u>

Step-by-step explanation:

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05

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Answer:

(E) 9

Step-by-step explanation:

A= 4

B= 5

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a+b=9= 4+5=9

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Step-by-step explanation:

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In a lottery game a player wins $1,000,000 with probability 0.0000005, wins $200,000 with probability0.000002, and wins $30,000
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Answer:

X_1 = 1000000 , P(X_1)= 0.0000005

X_2 = 200000 , P(X_2)= 0.000002

X_3 = 30000 , P(X_3)= 0.00001

X_4 = -2, P(X_4) = 1-0.0000005-0.000002-0.00001= 0.9999875

And then replacing we have this for the expected value:

E(X) = 1000000*0.0000005 +200000*0.000002+30000*0.00001 2*0.9999875

And after do the math we got:

E(X) = -0.79998

And then the expected value for this game is -0.79998 a negative result means NOT a win.

Step-by-step explanation:

For this case we can define the random variable X as the amount that we can win. We can find the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And we have the following probabilities and possible values given:

X_1 = 1000000 , P(X_1)= 0.0000005

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Answer:

P(X > 126) = 0.2119

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 110, \sigma = 20

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

Z = \frac{126 - 110}{20}

[tez]Z = 0.8[/tex]

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