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3 years ago

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Consider a sample with data values of 27, 24, 21, 16, 30, 33, 28, and 24. Compute the 20th, 25th, 65th, and 75th percentiles. 20

th percentile 20 Correct: Your answer is correct. 25th percentile 21 Incorrect: Your answer is incorrect. 65th percentile 27.5 Incorrect: Your answer is incorrect. 75th percentile 28 Incorrect: Your answer is incorrect.

1 answer:

densk [106]3 years ago

7 0

Answer:

$P_{20} = 20$ --- 20th percentile

$P_{25} = 21.75$ --- 25th percentile

$P_{65} = 27.85$ --- 65th percentile

$P_{75} = 29.5$ --- 75th percentile

Step-by-step explanation:

Given

27, 24, 21, 16, 30, 33, 28, and 24.

N = 8

First, arrange the data in ascending order:

Arranged data: 16, 21, 24, 24, 27, 28, 30, 33

Solving (a): The 20th percentile

This is calculated as:

$P_{20} = 20 * \frac{N +1}{100}$

$P_{20} = 20 * \frac{8 +1}{100}$

$P_{20} = 20 * \frac{9}{100}$

$P_{20} = \frac{20 * 9}{100}$

$P_{20} = \frac{180}{100}$

$P_{20} = 1.8th\ item$

This is then calculated as:

$P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)$

$P_{20} = 16 + 0.8*(21 - 16)$

$P_{20} = 16 + 0.8*5$

$P_{20} = 16 + 4$

$P_{20} = 20$

Solving (b): The 25th percentile

This is calculated as:

$P_{25} = 25 * \frac{N +1}{100}$

$P_{25} = 25 * \frac{8 +1}{100}$

$P_{25} = 25 * \frac{9}{100}$

$P_{25} = \frac{25 * 9}{100}$

$P_{25} = \frac{225}{100}$

$P_{25} = 2.25\ th$

This is then calculated as:

$P_{25} = 2nd\ item + 0.25(3rd\ item-2nd\ item)$

$P_{25} = 21 + 0.25(24-21)$

$P_{25} = 21 + 0.25(3)$

$P_{25} = 21 + 0.75$

$P_{25} = 21.75$

Solving (c): The 65th percentile

This is calculated as:

$P_{65} = 65 * \frac{N +1}{100}$

$P_{65} = 65 * \frac{8 +1}{100}$

$P_{65} = 65 * \frac{9}{100}$

$P_{65} = \frac{65 * 9}{100}$

$P_{65} = \frac{585}{100}$

$P_{65} = 5.85\th$

This is then calculated as:

$P_{65} = 5th + 0.85(6th - 5th)$

$P_{65} = 27 + 0.85(28 - 27)$

$P_{65} = 27 + 0.85(1)$

$P_{65} = 27 + 0.85$

$P_{65} = 27.85$

Solving (d): The 75th percentile

This is calculated as:

$P_{75} = 75 * \frac{N +1}{100}$

$P_{75} = 75 * \frac{8 +1}{100}$

$P_{75} = 75 * \frac{9}{100}$

$P_{75} = \frac{75 * 9}{100}$

$P_{75} = \frac{675}{100}$

$P_{75} = 6.75th$

This is then calculated as:

$P_{75} = 6th + 0.75(7th - 6th)$

$P_{75} = 28 + 0.75(30- 28)$

$P_{75} = 28 + 0.75(2)$

$P_{75} = 28 + 1.5$

$P_{75} = 29.5$

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Question 2:

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with P(A and B) beeing the probability of both events occurring simultaneously.

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B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

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$P ( Y)= \frac{ 4 }{ 16 }$

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P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} } = \frac{1}{2}$

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

$P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}$

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

B: Baseball League Championships won, beeing

$P ( B ) = \frac{ 6 }{16}$

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

$P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}$

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

$P ( Y)= \frac{ 4 }{ 16 }$

Z: Championships won by the 13 - 15 years old, beeing

$P ( Z)= \frac{ 1 }{ 16 }$

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By definition,

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