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IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X= IQ of an individual.(a) Find the z-score for an IQ of 124, rounded to three decimal places. (b) Find the probability that the person has an IQ greater than 124.

User Skbrhmn
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In this problem we have a normal distribution with:

• μ = mean = 100,

,

• σ = standard deviation = 15.

(a) The z-score for a value x = 124 is:


z=(x-\mu)/(\sigma)=(124-100)/(15)=(24)/(15)=1.6.

(b) We must compute the probability that x is greater than 124, which is equivalent to the probability of having z greater than 1.6. Using a table of z-score, we find that:


P(x>124)=P(z>1.6)=0.0548.

Answer

• (a), z = 1.6

,

• (b), P(x > 124) = 0.0548

User Flolo
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