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With reference to the distribution of IQ scores, according to the 68-95-99.7 rule, what is the probability that a person selected at random has an IQ greater than 100?

Answer :

The IQ score distribution follows a normal curve and is distributed with a mean of 100 and a standard deviation of 15. The 68-95-99.7 rule states that approximately 68% of the population falls within one standard deviation of the mean, 95% falls within two standard deviations of the mean, and 99.7% falls within three standard deviations of the mean.

To find the probability that a person selected at random has an IQ greater than 100, we need to calculate the z-score first. The z-score formula is given by:
z = (X - μ) / σ
where X is the IQ score, μ is the mean, and σ is the standard deviation.

Substituting the given values, we get:
z = (100 - 100) / 15
z = 0

A z-score of 0 means that the IQ score is equal to the mean. Since we want to find the probability of a person having an IQ score greater than 100, we need to find the area under the normal curve to the right of z = 0. Using a standard normal distribution table or a calculator, we can find this area to be approximately 0.5 or 50%. Therefore, the probability that a person selected at random has an IQ greater than 100 is 50%.

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