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valentinak56 [21]
3 years ago
5

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

Mathematics
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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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th
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The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

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

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

And the probability of each is:

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

P(1) = \frac{2}{6}

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

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

P(2) = \frac{3}{6}

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

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

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

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

P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}

Take LCM

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>

7 0
2 years ago
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