Suppose That IQ Scores In One Region Are Normally Distribute

Suppose That Iq Scores In One Region Are Normally Distributed With

Suppose that IQ scores in one region are normally distributed with a standard deviation of 18. Suppose also that exactly 55% of the individuals from this region have IQ scores greater than 100. What is the mean IQ score for this region? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Sample Paper For Above instruction

Understanding the distribution of IQ scores within a population is fundamental in psychological and educational assessments. When IQ scores are normally distributed, the mean and standard deviation define the entire distribution, enabling us to determine probabilities and specific percentile scores. This paper seeks to determine the mean IQ score for a given region where the IQ scores follow a normal distribution with a known standard deviation and a specified proportion of individuals scoring above a certain IQ value.

Given that the IQ scores in this region are normally distributed with a standard deviation (σ) of 18, and that exactly 55% of individuals score above an IQ of 100, our goal is to find the mean IQ score (μ) of this distribution. The problem provides two key pieces of information: the proportion of individuals exceeding a specific IQ score and the standard deviation of the scores, which together allow us to leverage properties of the standard normal distribution.

To approach this problem, we recognize that the probability P(X > 100) = 0.55. Repartitioning the problem in terms of the standard normal variable Z, where Z = (X - μ) / σ, the probability becomes P(Z > (100 - μ) / 18). Since P(Z > z) = 0.55, this corresponds to the area to the right of z in the standard normal distribution. By symmetry, P(Z

Using standard normal distribution tables or a statistical software, the z-score corresponding to a cumulative probability of 0.45 is approximately -0.1257. Therefore, we have:

z = (100 - μ) / 18 = -0.1257

Solving for μ gives:

μ = 100 - (z 18) = 100 - (-0.1257 18) = 100 + 2.2626 = 102.3

Thus, the mean IQ score for this region is approximately 102.3. This means that, despite a standard deviation of 18, the population's IQ scores tend to cluster around a mean of about 102.3, consistent with the observed proportion exceeding 100.

In conclusion, determining the mean of a normally distributed IQ score dataset with known standard deviation and percentile information involves converting the given probability into a z-score, then solving for the mean. This process illustrates the utility of the properties of the normal distribution in psychological testing and population studies, enabling precise estimation of central tendencies based on percentile data.

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