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Some IQ tests are standardized to a Normal model, with a mean of 100 and a standard deviation of \(16 .\) a) Draw the model for these IQ scores. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b) In what interval would you expect the central \(95 \%\) of IQ scores to be found? c) About what percent of people should have IQ scores above \(116 ?\) d) About what percent of people should have IQ scores between 68 and 84 ? e) About what percent of people should have IQ scores above \(132 ?\)

Short Answer

Expert verified
a) Use the normal curve with mean 100 and mark \( \,1\sigma, \,2\sigma, \,3\sigma\) intervals. b) 68 to 132. c) About 15.87%. d) About 18.26%. e) About 2.28%.

Step by step solution

01

Understand the Normal Distribution

The given IQ scores follow a normal distribution model with a mean \( \mu \) of 100 and a standard deviation \( \sigma \) of 16. The normal distribution is symmetric around the mean.
02

Visual Representation of the Model

Draw a bell-shaped normal curve centered at 100. Use the \(68-95-99.7\) rule to mark standard deviations: 68% of data falls within \( \mu \pm \sigma \), 95% within \( \mu \pm 2\sigma \), and 99.7% within \( \mu \pm 3\sigma \).
03

Calculating the 95% Interval

The 95% interval is within \(2\sigma\) of the mean. Calculate the interval as \(100 - 2(16) \) to \(100 + 2(16)\), which is 68 to 132.
04

Percent of People with IQ above 116

Convert the IQ score 116 into the z-score using \( z = \frac{116 - 100}{16} = 1\). From the standard normal distribution table, \( P(Z > 1) \approx 0.1587 \) or 15.87%.
05

Percent of People with IQ between 68 and 84

Convert both IQ scores into z-scores: for 68, \( z = \frac{68 - 100}{16} = -2\) and for 84, \( z = \frac{84 - 100}{16} = -1\). From the standard normal distribution, \( P(-2 < Z < -1) \approx 0.3413 - 0.1587 = 0.1826 \) or 18.26%.
06

Percent of People with IQ above 132

Convert the IQ score 132 into the z-score: \( z = \frac{132 - 100}{16} = 2\). The probability \( P(Z > 2) \approx 0.0228 \) or 2.28%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
The standard deviation is a measure that tells us how much the numbers in a data set typically differ from the average (mean) of the set. In the context of a normal distribution, it plays a key role in understanding the spread of the distribution.
In our IQ test example, the standard deviation is 16. This means that most IQ scores tend to hover within 16 points of the average IQ score of 100. Having a smaller standard deviation would mean that the IQ scores are closer to the mean, reflecting lesser individual variation, whereas a larger value would mean greater variety in IQ scores.
  • It's used to calculate ranges of values where different portions of the data are expected.
  • It helps determine how uncommon a particular score is relative to others.
In normal distributions, key intervals are measured using standard deviations from the mean, helping us to quickly see where most of the data lies and make predictions.
Z-Score
A z-score tells us how many standard deviations away a value is from the mean. It converts a particular data point into a standardized form, providing a clear sense of its relative position in the distribution.
To calculate a z-score, subtract the mean from the point and divide the result by the standard deviation. For example, an IQ score of 116 translates to a z-score:\[z = \frac{116 - 100}{16} = 1\]This calculation shows that 116 is one standard deviation above the mean.
  • Z-scores help determine probabilities for values in a normal distribution.
  • They allow comparison of data points across different distributions by standardizing them.
Using z-scores, we can look up corresponding probabilities in a standard normal distribution table, or use statistical software to find the proportion of data above, below, or between certain values.
68-95-99.7 Rule
The 68-95-99.7 rule, also known as the empirical rule, helps us understand the distribution of data when plotted on a normal curve. It gives us a quick overview of where the majority of data lies relative to the mean.
According to the rule:
  • About 68% of values fall within one standard deviation from the mean.
  • Roughly 95% are within two standard deviations.
  • And nearly 99.7% are within three standard deviations.
Applying this to our IQ distribution, we see that: - 68% of IQ scores fall between 84 (100 - 16) and 116 (100 + 16). - 95% range from 68 (100 - 2×16) to 132 (100 + 2×16). - 99.7% lie between 52 (100 - 3×16) and 148 (100 + 3×16). This rule is handy for quickly estimating which values are common or rare in any normal distribution.

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Most popular questions from this chapter

A company selling clothing on the Internet reports that the packages it ships have a median weight of 68 ounces and an IQR of 40 ounces. a) The company plans to include a sales flyer weighing 4 ounces in each package. What will the new median and IQR be? b) If the company recorded the shipping weights of these new packages in pounds instead of ounces, what would the median and IQR be? \((1 \mathrm{lb} .=16 \mathrm{oz}\).)

Companies that design furniture for elementary school classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children can be described by a Normal model with a mean of \(38.2\) inches and standard deviation of \(1.8\) inches. a) What fraction of kindergarten kids should the company expect to be less than 3 feet tall? b) In what height interval should the company expect to find the middle \(80 \%\) of kindergarteners? c) At least how tall are the biggest \(10 \%\) of kindergarteners?

Based on the model \(N(1152,84)\) describing Angus steer weights, what are the cutoff values for a) the highest \(10 \%\) of the weights? b) the lowest \(20 \%\) of the weights? c) the middle \(40 \%\) of the weights?

Here are the summary statistics for the weekly payroll of a small company: lowest salary \(=\$ 300\), mean salary \(=\$ 700\), median \(=\$ 500\), range \(=\$ 1200, \mathrm{IQR}=\) \(\$ 600\), first quartile \(=\$ 350\), standard deviation \(=\$ 400\). a) Do you think the distribution of salaries is symmetric, skewed to the left, or skewed to the right? Explain why. b) Between what two values are the middle \(50 \%\) of the salaries found? c) Suppose business has been good and the company gives every employee a \(\$ 50\) raise. Tell the new value of each of the summary statistics. d) Instead, suppose the company gives each employee a \(10 \%\) raise. Tell the new value of each of the summary statistics.

Recall that the beef cattle described in Exercise 17 had a mean weight of 1152 pounds, with a standard deviation of 84 pounds. a) Cattle buyers hope that yearling Angus steers will weigh at least 1000 pounds. To see how much over (or under) that goal the cattle are, we could subtract 1000 pounds from all the weights. What would the new mean and standard deviation be? b) Suppose such cattle sell at auction for 40 cents a pound. Find the mean and standard deviation of the sale prices for all the steers.

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