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Sonbull [250]
3 years ago
11

Assume that adults have it scores that are normally distributed with a mean of 100 standard deviation of 15 find probability tha

t randomly selected adult has an Iq between 89 and 111
Mathematics
1 answer:
exis [7]3 years ago
4 0

Answer:

0.5346

Step-by-step explanation:

Find the z-scores.

z = (x − μ) / σ

z₁ = (89 − 100) / 15

z₁ = -0.73

z₂ = (111 − 100) / 15

z₂ = 0.73

Find the probability.

P(-0.73 < Z < 0.73)

= P(Z < 0.73) − P(Z < -0.73)

= 0.7673 − 0.2327

= 0.5346

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First order the data, ascendingly:

10, 17, 18, 18, 18, 20, 20, 25, 25, 30, 30

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For an even number of values: The number of values / 2

For an odd number of values: The number of values + 1 / 2

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11 + 1 / 2

12 / 2 = 6

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10, 17, 18, 18, 18, 20, 20, 25, 25, 30, 30

The median is 20.

<em>Other method:</em>

<em>You don't need this step.</em>

Using the values lower than the median, exclusive, find the median of the lower values;

For an odd number of values: <u>The number of values + 1 / 2</u>

10, 17, 18, 18, 18

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10, 17, 18, 18, 18

<em>Other method:</em>

<em>For an odd number of values: </em><u><em>The number of values + 1 / 4</em></u>

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<em>10, 17, </em><em>18,</em><em> 18, 18, 20, 20, 25, 25, 30, 30</em>

<em />

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For an odd number of values: <u>The number of values + 1 / 2</u>

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<em>For an odd number of values: </em><u><em>(The number of values + 1 / 4) * 3</em></u>

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<em>10, 17, 18</em><em>,</em><em> 18, 18, 20, 20, 25, </em><em>25</em><em>, 30, 30</em>

Your interquartile range is 3 - 25

25 - 3 = 21

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