To solve this problem, we need to use the properties of the standard normal distribution.

1. First, we need to find the z-score that corresponds to the 52nd percentile of the standard normal distribution. This is because the problem tells us that 52% of individuals have an IQ score greater than 100. In terms of the standard normal distribution, this means that 100 is the 52nd percentile.

2. We can find this z-score using a standard normal distribution table or a calculator with a built-in normal distribution function. The z-score that corresponds to the 52nd percentile is approximately 0.05.

3. Once we have the z-score, we can use the formula for converting z-scores to raw scores in a normal distribution, which is X = μ + Zσ. In this formula, X is the raw score, μ is the mean, Z is the z-score, and σ is the standard deviation.

4. We know that X = 100 (the IQ score at the 52nd percentile), Z = 0.05 (the z-score for the 52nd percentile), and σ = 16 (the standard deviation of the IQ scores). We can plug these values into the formula and solve for μ (the mean IQ score).

5. The equation becomes 100 = μ + 0.05*16. Solving for μ, we get μ = 100 - 0.05*16 = 99.2.

So, the mean IQ score for this region is approximately 99.2.

Question

To solve this problem, we need to use the properties of the standard normal distribution.

1. First, we need to find the z-score that corresponds to the 52nd percentile of the standard normal distribution. This is because the problem tells us that 52% of individuals have an IQ score greater than 100. In terms of the standard normal distribution, this means that 100 is the 52nd percentile.

2. We can find this z-score using a standard normal distribution table or a calculator with a built-in normal distribution function. The z-score that corresponds to the 52nd percentile is approximately 0.05.

3. Once we have the z-score, we can use the formula for converting z-scores to raw scores in a normal distribution, which is X = μ + Zσ. In this formula, X is the raw score, μ is the mean, Z is the z-score, and σ is the standard deviation.

4. We know that X = 100 (the IQ score at the 52nd percentile), Z = 0.05 (the z-score for the 52nd percentile), and σ = 16 (the standard deviation of the IQ scores). We can plug these values into the formula and solve for μ (the mean IQ score).

5. The equation becomes 100 = μ + 0.05*16. Solving for μ, we get μ = 100 - 0.05*16 = 99.2.

So, the mean IQ score for this region is approximately 99.2.

Knowee AI · Accepted Answer

To solve this problem, we need to use the properties of the standard normal distribution.

1. First, we need to find the z-score that corresponds to the 52nd percentile of the standard normal distribution. This is because the problem tells us that 52% of individuals have an IQ score greater than 100. In terms of the standard normal distribution, this means that 100 is the 52nd percentile.

2. We can find this z-score using a standard normal distribution table or a calculator with a built-in normal distribution function. The z-score that corresponds to the 52nd percentile is approximately 0.05.

3. Once we have the z-score, we can use the formula for converting z-scores to raw scores in a normal distribution, which is X = μ + Zσ. In this formula, X is the raw score, μ is the mean, Z is the z-score, and σ is the standard deviation.

4. We know that X = 100 (the IQ score at the 52nd percentile), Z = 0.05 (the z-score for the 52nd percentile), and σ = 16 (the standard deviation of the IQ scores). We can plug these values into the formula and solve for μ (the mean IQ score).

5. The equation becomes 100 = μ + 0.05*16. Solving for μ, we get μ = 100 - 0.05*16 = 99.2.

So, the mean IQ score for this region is approximately 99.2.

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