Question

In the Normal model \(N(100,16)\), what cutoff value bounds a) the highest \(5 \%\) of all IQs? b) the lowest \(30 \%\) of the IQs? c) the middle \(80 \%\) of the IQs?

Short answer

a) 126.32; b) 91.62; c) 79.52 to 120.48

Step-by-step solution

Understanding the Normal Distribution

The problem specifies a normal distribution with a mean (\(\mu\)) of 100 and a standard deviation (\(\sigma\)) of 16. We denote this distribution as \(N(100, 16)\). Understanding the properties of a normal distribution is crucial, as it helps us utilize the standard normal distribution tables (z-tables) or software to find proportionality cutoffs.

Transform to Standard Normal Distribution

To find the cutoff values, we use the standard normal distribution where the mean is 0 and standard deviation is 1. The formula to convert a score \(X\) from a normal distribution \(N(\mu, \sigma)\) to the standard normal distribution \(Z\) is presented as:\[ Z = \frac{X - \mu}{\sigma} \].

Finding Cutoff for Highest 5% of IQs

To find the cutoff for the highest 5% of IQs, we find the corresponding z-score for a 95% percentile (since the highest 5% lies above this point). Using a z-table or calculator, we find the z-score that corresponds to 95%. Let's denote this z-value as \(z_{0.95}\). Assuming \(z_{0.95} \approx 1.645\), we convert back to the IQ scale:\[ X_{high} = \mu + (z_{0.95} \times \sigma) = 100 + (1.645 \times 16) \approx 126.32 \].

Finding Cutoff for Lowest 30% of IQs

For the lowest 30%, we look for the z-score that corresponds to the 30th percentile. Denote this z-value as \(z_{0.30}\). Assuming \(z_{0.30}\approx -0.524\), we convert back to the IQ scale:\[ X_{low} = \mu + (z_{0.30} \times \sigma) = 100 + (-0.524 \times 16) \approx 91.62 \].

Finding Cutoff for Middle 80% of IQs

The middle 80% means excluding 10% on both extremes, requiring the 10th and 90th percentiles. Find z-scores \(z_{0.10}\) and \(z_{0.90}\) assuming they are approximately -1.28 and 1.28.Convert these to IQ using:\[ X_{10\%} = \mu + (z_{0.10} \times \sigma) = 100 + (-1.28 \times 16) = 79.52 \].\[ X_{90\%} = \mu + (z_{0.90} \times \sigma) = 100 + (1.28 \times 16) = 120.48 \].

Key concepts

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution. In this case, it has a mean of 0 and a standard deviation of 1. This simplifies many calculations and makes it easier to work with probabilities and percentiles. By converting a normal distribution to the standard normal distribution, we can use tools like the Z-table to find statistical probabilities and cutoff values more efficiently. How do we go from a normal distribution to a standard normal distribution? We use the formula: \[ Z = \frac{X - \mu}{\sigma} \]This converts a value from your normal distribution, where your mean is \( \mu \) and your standard deviation is \( \sigma \), to our standard normal distribution, where the mean is 0 and the standard deviation is 1. This transformation simplifies calculations, especially when dealing with cutoff values, as it allows us to align different distributions into a common framework. In our IQ problem, we're working with a normal model \(N(100,16)\). To find specific percentile cutoffs, such as the highest 5% or the lowest 30%, we use the standard normal distribution.

Z-scores

Z-scores are a key concept when dealing with normal distributions. A Z-score tells you how many standard deviations a value \(X\) is from the mean \(\mu\). Calculating Z-scores allows you to understand where a value fits within a distribution and compare it to other distributions. For any given value \(X\), the Z-score is calculated using:\[ Z = \frac{X - \mu}{\sigma} \]By standardizing scores, you can interpret the data more easily. For example, how does a score of 126 compare to the mean of 100 when the standard deviation is 16? This becomes clear through Z-scores.From the exercise:- To find the top 5% cutoff, we use the Z-score corresponding to the 95th percentile, approximately \(1.645\).- For the lowest 30%, we find the Z-score of the 30th percentile, around \(-0.524\).These Z-scores can be translated back into our original IQ values by rearranging our formula to \(X = \mu + Z \times \sigma\). Understanding this concept helps you quickly identify how likely a particular score is based on its distance from the mean.

Percentiles

Percentiles give us a way to understand positions in a probability distribution. A percentile indicates the value below which a given percentage of observations in a group fall. For example, if a student scores in the 90th percentile, their score is higher than 90% of all the scores.In a normal distribution like IQ scores, percentiles can be identified using Z-scores:- To find the top 5% cutoff IQ, we used the 95th percentile Z-score (\(1.645\)). This corresponds to the top 5% ranked scores.- For the lowest 30%, it's the 30th percentile score, which means 70% are above.- For the middle 80%, we focus on the 10th to 90th percentile range.The beauty of percentiles is they provide an intuitive way to evaluate an individual's position compared to the rest without needing to understand complex statistical concepts. They essentially translate the Z-scores into more understandable rankings.