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Assume that adults have IQ scores that are normally distributed with a mean of 98.2 and a standard deviation of 19.6. Find the first quartile \(Q_1\), which is the lowest score separating the bottom 25% from the top 75%. (Hint: Draw a graph.)

The first quartile is:

Type an integer or decimal rounded to one decimal place as needed.

Answer :

Final answer:

The first quartile (Q1), which separates the bottom 25% from the top 75% of scores in a normal distribution with a mean of 98.2 and a standard deviation of 196, is approximately 71.2.

Explanation:

The question can be solved using the concept of the Normal Distribution in Statistics. We are given the mean, standard deviation and need to find the first quartile (Q1), i.e., the score which separates the bottom 25% from the top 75% of the data.

First and foremost, we need to find the associated z-score with the first quartile. The z-score for the first quartile can be looked up in a standard distribution table or computed programmatically, and is approximately -0.67.

After that, we use the formula X = μ + Zσ, where X represents the score we are looking for, μ is the mean, Z is the z-score, and σ is the standard deviation. Substituting the given values, we find X = 98.2 + (-0.67 * 196) ≈ 71.2.

Therefore, the first quartile, Q1, is 71.2.

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