Question 425598: Bob takes an online IQ test and finds that his IQ according to the test is 134. Assuming that the mean IQ is 100, the standard deviation is 15, and the distribution of IQ scores is normal, what proportion of the population would score higher than Bob? Lower than Bob?
According to Steinberg (2011), The Z-score is found by assuming that the null hypothesis is true. Therefore, the formula to figure out the z score would be z=(X-M)/SD. Z=(134-100)/15. 34/15=0.44.
The z score of 0.44 would be used to determine the p value.
According to Appendix A, the p score would be .670
Since, the z score is positive, it would be subtracted by 1.000- 0.44=.56.
Therefore the area between the mean and z would be 0.44 and
the area beyond mean and z is .560.
There would be 44% who are lower than Bob
There would be 56% who are higher than Bob
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Bob takes an online IQ test and finds that his IQ according to the test is 134. Assuming that the mean IQ is 100, the standard deviation is 15, and the distribution of IQ scores is normal, what proportion of the population would score higher than Bob? Lower than Bob?
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z(134) = (134-100)/15 = 2.2667
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P(x > 134) = P(z > 2.2667) = 0.0117
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P(x < 134) = P(z < 2.2667) = 0.9883
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Cheers,
Stan H.
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