IQ scores are normally distributed with a mean of 110 and a standard deviation of 16. Find the probability a randomly selected person has an IQ score greater than 115.
Here, μ = 110, σ = 16 and x = 115. We need to compute P(X >= 115). The corresponding z-value is calculated using Central Limit Theorem
z = (x - μ)/σ
z = (115 - 110)/16 = 0.31
Therefore,
P(X >= 115) = P(z <= (115 - 110)/16)
= P(z >= 0.31)
= 1 - 0.6217 = 0.3783
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