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Ket [755]
3 years ago
9

Suppose the test scores on an exam show a normal distribution with a mean of 82 and a standard deviation of

Mathematics
1 answer:
Zanzabum3 years ago
5 0

Answer:

a) Between 74 and 90

b) 81.86% of the scores are between 78 and 90

Step-by-step explanation:

a) According to the Empirical Rule, in a normal distribution, 95% of the data are within 2 standard deviations of the mean. Therefore, the range that 95% of the scores fall in is between 82-4(2)=74 and 82+4(2)=90.

b) The percent of scores that are between 78 and 90 is: normalcdf(lower,upper,μ,σ) = normalcdf(78,90,82,4) = 0.8185946784 ≈ 0.8186, or about 81.86% of the scores.

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Answer:

Step-by-step explanation:

Confidence interval for the difference in the two proportions is written as

Difference in sample proportions ± margin of error

Sample proportion, p= x/n

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For the men,

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Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]

To determine the z score, we subtract the confidence level from 100% to get α

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3 years ago
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F(x) = square root 3x+7 , g(x) = square root 3x-7<br><br> Find (f + g)(x).
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Then go ahead and square both sides to remove the radical.
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Now if you kept trying to isolate x, you find that both sides will just cancel each other out and you are left with,
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