To find the proportion of the population with an IQ above 146, we first need to calculate the z-score for an IQ of 146. The z-score is a measure of how many standard deviations an element is from the mean.

The formula for the z-score is (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.

So, for an IQ of 146, the z-score would be (146 - 100) / 15 = 3.07.

In a normal distribution, about 99.9% of the data falls below 3 standard deviations above the mean. This is a known value from the standard normal distribution table.

So, if 99.9% of people have an IQ below 146, then 100% - 99.9% = 0.1% of people have an IQ above 146.

Expressed as a decimal to three decimal places, this is 0.001.

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To find the proportion of the population with an IQ above 146, we first need to calculate the z-score for an IQ of 146. The z-score is a measure of how many standard deviations an element is from the mean.

The formula for the z-score is (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.

So, for an IQ of 146, the z-score would be (146 - 100) / 15 = 3.07.

In a normal distribution, about 99.9% of the data falls below 3 standard deviations above the mean. This is a known value from the standard normal distribution table.

So, if 99.9% of people have an IQ below 146, then 100% - 99.9% = 0.1% of people have an IQ above 146.

Expressed as a decimal to three decimal places, this is 0.001.

Knowee AI · Accepted Answer

To find the proportion of the population with an IQ above 146, we first need to calculate the z-score for an IQ of 146. The z-score is a measure of how many standard deviations an element is from the mean.

The formula for the z-score is (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.

So, for an IQ of 146, the z-score would be (146 - 100) / 15 = 3.07.

In a normal distribution, about 99.9% of the data falls below 3 standard deviations above the mean. This is a known value from the standard normal distribution table.

So, if 99.9% of people have an IQ below 146, then 100% - 99.9% = 0.1% of people have an IQ above 146.

Expressed as a decimal to three decimal places, this is 0.001.

IQ scores are normally distributed with mean 100 and standard deviation 15.2. What proportion of the population would have IQ above Marvin's score of 146? (express as a decimal, to 3 decimal places)

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