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Suppose that IQ scores in one region are normally distributed with a standard deviation of 16. Suppose also that exactly 52% of the individuals from this region have IQ scores of greater than 100 (and that 48% do not). What is the mean IQ score for this region

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Answer: 100.8

Step-by-step explanation:

Given that:

Standard deviation (σ) = 16

X > 100

From the z table :

Zscore = P(X > 100) = - 0.05

Zscore = - 0.05

Thus ;

Zscore = (x - m) / σ

x = 100

-0.05 = ( 100 - m) / 16

-0.05 * 16 = 100 - m

- 0.8 = 100 - m

m = 100 + 0.8

Mean = 100. 8

Mean score = 100.8

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