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
1 answer:
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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