Question

Based on the Normal model \(N(100,16)\) describing IQ scores, what percent of people's IQs would you expect to be a) over \(80 ?\) b) under \(90 ?\) c) between 112 and \(132 ?\)

Short answer

a) 100% over 80. b) 0.62% under 90. c) 0.13% between 112 and 132.

Step-by-step solution

Identify the Normal Distribution

We are given a normal distribution for IQ scores described by the model \(N(100,16)\), where the mean \(\mu = 100\) and the variance \(\sigma^2 = 16\). This implies the standard deviation \(\sigma = 4\).

Convert Score to Z-Score for Over 80

To find the percentage of people with an IQ over 80, we first convert the IQ score of 80 to a Z-score using the formula \(Z = \frac{X - \mu}{\sigma}\). Thus, \(Z = \frac{80 - 100}{4} = -5\).

Find Probability from Z-Table for Over 80

Using the Z-table, a Z-score of -5 is extremely low, practically resulting in a probability of nearly 0 (or 0.0000) for people scoring 80 or below. Consequently, the probability of scoring over 80 is approximately 1 (or 100%).

Convert Score to Z-Score for Under 90

To find the percentage of people with an IQ under 90, convert the IQ score of 90 to a Z-score using the formula \(Z = \frac{X - \mu}{\sigma}\). Thus, \(Z = \frac{90 - 100}{4} = -2.5\).

Find Probability from Z-Table for Under 90

Using the Z-table, a Z-score of -2.5 corresponds to a cumulative probability of approximately 0.0062. This means about 0.62% of people have an IQ under 90.

Convert Scores to Z-Scores for Between 112 and 132

To find the percentage of people with an IQ between 112 and 132, convert these scores to Z-scores. For 112, \(Z = \frac{112 - 100}{4} = 3\), and for 132, \(Z = \frac{132 - 100}{4} = 8\).

Find Probability from Z-Table for Between 112 and 132

Using the Z-table, a score of \(Z = 3\) gives a cumulative probability of roughly 0.9987. Since a Z-score of 8 is far beyond usual tables, it approximates to 1, indicating a nearly full range from Z = 3 to Z = 8. Hence, the probability is approximately 0.0013 or 0.13% for having an IQ between 112 and 132.

Key concepts

Z-Score

The Z-Score is a powerful statistical tool used to determine how far away a data point is from the mean in the context of a standard normal distribution.

It is calculated using the formula: \[Z = \frac{X - \mu}{\sigma}\]Where:

• \(Z\) is the Z-score.
• \(X\) is the data point or score you are examining, such as an IQ score.
• \(\mu\) is the mean of the distribution.
• \(\sigma\) is the standard deviation of the distribution.

The Z-score shows the number of standard deviations a particular score is away from the average. For example, a Z-score of -5 suggests the IQ is 5 standard deviations below the mean, while a Z-score of 3 places the IQ 3 standard deviations above the mean.
In our example, Z-scores help us determine probabilities using the Z-table, a statistical tool listing probabilities that a standard normal random variable is less than or equal to a given value.

Probability

Probability is the measure of the likelihood or chance of a particular event occurring. In statistical terms, it's a value between 0 (impossible event) and 1 (certain event), that signifies how often you would expect to see the given event in repeated trials.

In the context of the normal distribution and IQ scores, probabilities inform us about the percentage of individuals likely to fall within a certain range of scores.

For example:

• A Z-score with a very low value like -5 has a probability near 0, meaning almost nobody scores 80 or below when the mean IQ is 100.
• A Z-score of -2.5 corresponds to a probability of around 0.0062, indicating about 0.62% have IQs under 90.
• Probabilities for a range of scores, like between 112 and 132, explain the very small percent (here, just 0.13%) of people falling within this higher bracket.

This is how probabilities provide insight into the distribution of IQ scores within a population.

IQ Scores

IQ Scores, or Intelligence Quotient Scores, are standardized measures used to assess human intelligence. These scores are typically modeled using a normal distribution, with specific mean and standard deviation values that represent the average and spread of intelligence levels across the population, respectively.

In our example, the IQ scores are modeled by the normal distribution \(N(100,16)\), where:

• \(\mu = 100\): This is the mean, representing the average IQ score.
• \(\sigma^2 = 16\): This is the variance, with the square root (\(\sigma = 4\)) being the standard deviation.

This model signifies that most people have an IQ score close to 100, with fewer individuals having significantly higher or lower scores.
Understanding the IQ score distribution allows us to use statistics to estimate how common a certain IQ score is among the population. Identifying outliers or extreme values becomes easier with this model; IQs well above or below average show their rarity in this context.

Standard Deviation

Standard Deviation is a crucial measure in statistics that indicates how spread out the numbers in a set of data are. It is the square root of the variance, providing an understanding of how much variation or dispersion exists from the average (mean).

In a normal distribution, approximately 68% of data fall within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

• For IQ scores modeled as \(N(100,16)\), the standard deviation \(\sigma = 4\).
• This indicates that a majority of IQ scores lie close to 100, within 4 points in either direction for roughly two-thirds of the population.
• Less common are scores several standard deviations away, confirming either unusually high or low intelligence.

This concept is central when working with Z-scores, as it allows conversion of raw data into standardized scores that describe their distance from the mean in terms of standard deviations. This helps in determining probabilities and understanding the distribution of scores in a population.