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valentinak56 [21]

3 years ago

5

PLEASE HELP me ANSWER these!!! My grade depends on it!!

1 answer:

iren [92.7K]3 years ago

8 0

M (Mean) = 20,500;
SD (Standard Deviation) = 55
1 ) 20,555 = 20,500 + 55 = M + 1 SD
Answer: 50% + 34.1% = 84.1%
2 ) 20.610 = 20,000 + 110 = M + 2 SD
Answer: 10% - ( 50% + 34.1% + 13.6% ) = 100% - 97.7% = 2.3%
3 ) 20,445 = 20,500 - 55 = M - 1 SD
Answer: 50% - 34.1% = 15.9%

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Analysing the question one statement at a time.

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$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

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