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Fiesta28 [93]
4 years ago
13

what are the minimume, first quartile, median, third quartile, and maximum of the data set? 2 6 12 8 3 9 14 20

Mathematics
2 answers:
Gre4nikov [31]4 years ago
4 0
To get the indicated values we re-arrange the set of numbers in ascending order:
<span>2 6 12 8 3 9 14 20
rearranging gives us:
2,3,6,8,9,12,14,20
Minimum-2
First quartile-5.25
Median-8.5
Third quartile-12.50
Maximum-20

</span>
Blababa [14]4 years ago
4 0

Answer:

Given data set is : 2 6 12 8 3 9 14 20

We shall re-arrange the set of numbers in ascending order to get a clear concept:

2 3 6 8 9 12 14 20

1. Minimum: The minimum is the least value of the data set that is 2.

First quartile: The first quartile value is the median for the numbers before the median. 2 3 6 8 the value is \frac{2+3+6+8}{4}= 4.75

Median: The average of the center two numbers. \frac{8+9}{2}=8.5

Third quartile: The third quartile is the same as first but take last 4 values, 9 12 14 20 the value is \frac{9+12+14+20}{4}= 13.75

Maximum: The highest value number in the data set which is 20.

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Answer:25th Percentile: 5

50th Percentile: 30

75th Percentile: 34

Interquartile Range: 29

Step-by-step explanation:

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A right triangle has one side that measures 4 in. The angle opposite that side measures 80°. What is the length of the hypotenus
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3 years ago
207 students went to Rye Playland, Five buses were filled and 7 students traveled in cars,
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3 years ago
I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know
azamat

By definition of conditional probability,

P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}

=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}

Assuming a standard 6-sided fair die,

  • if X_1=4, then X_1+X_2=7 means X_2=3; otherwise,
  • if X_2=4, then X_1=3.

Both outcomes are mutually exclusive with probability \frac1{36} each, hence total probability \frac2{36}=\frac1{18}.

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs \frac6{36}=\frac16 of the time.

Then the probability we want is

P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13

6 0
3 years ago
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