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Assume a normal distribution with a mean of 70 and a standard deviation of 12 . What limits would include the middle \(65 \%\) of the cases?

Short Answer

Expert verified
The middle 65% of cases are between 58.792 and 81.208.

Step by step solution

01

Understand the Given Information

We have a normal distribution with a mean (\( \mu \)) of 70 and a standard deviation (\( \sigma \)) of 12. We need to find the limits that would include the middle 65% of the distribution.
02

Convert Percentage to Z-Scores

The middle 65% of the distribution means we are excluding 35%, or 17.5% from each tail. Using a Z-score table or calculator, identify the Z-scores corresponding to the lower 17.5% and the upper 82.5%. These are approximately -0.934 and 0.934, respectively.
03

Calculate the Raw Scores

Convert the Z-scores into raw scores using the formula: \[ X = \mu + Z \times \sigma \]. For the lower limit, \( Z = -0.934 \): \( X = 70 + (-0.934) \times 12 \). For the upper limit, \( Z = 0.934 \): \( X = 70 + 0.934 \times 12 \).
04

Compute the Results

Calculate the raw scores for both Z-values: For the lower limit: \( X = 70 - 11.208 \approx 58.792 \).For the upper limit: \( X = 70 + 11.208 \approx 81.208 \).
05

Conclusion

The middle 65% of the distribution lies between the raw scores of approximately 58.792 and 81.208.

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Key Concepts

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

Mean and Standard Deviation
In the world of statistics, understanding the mean and standard deviation is essential. A **mean**, often symbolized by \( \mu \), is the average value of a set of numbers. It provides a central point for the data, giving an idea of where most values lie.

When a distribution is normal, its shape resembles a bell curve. This curve is symmetric, with the mean at its center. **Standard deviation**, represented by \( \sigma \), measures how spread out the values are around the mean. A smaller standard deviation means data is tightly packed around the mean. Conversely, a larger standard deviation indicates data is more spread out.

In our example, the mean is 70, and the standard deviation is 12, giving insight into the data spread and central tendency. These two parameters help define the behavior of the normal distribution.
Z-Scores
**Z-scores** are a handy statistical tool that helps us understand how far a value is from the mean, in terms of standard deviations. Calculating a Z-score involves subtracting the mean from a value and then dividing by the standard deviation. It transforms our data point into a common scale.

This makes it easier to determine where it stands in the context of the entire dataset. The formula for converting a data point \( X \) into a Z-score is:
  • \( Z = \frac{X - \mu}{\sigma} \)
In a normal distribution, Z-scores allow comparison of the positions of different values relative to the spread of all data. They play a vital role when calculating percentiles and understanding relative standings within data.
Percentile Calculation
**Percentiles** help us understand the relative standing of a value within a distribution. The percentile of a score is the percentage of values in a distribution that are below it. For instance, if a value is at the 70th percentile, it means 70% of the values are below it.

In our scenario, the middle 65% of the data is considered. This means 17.5% of the data lies in each tail of the distribution after removing the middle part.

Using a Z-score table or calculator, we determine Z-scores corresponding to the 17.5th and 82.5th percentiles for a normal distribution. These Z-scores guide us in finding the raw scores within the spread that cover the desired middle percentage.
Raw Score Conversion
Converting **Z-scores** back into **raw scores** provides practical data values we can interpret within a specific context. The transformation is done using a conversion formula which relates Z-scores, the mean, and standard deviation.

For a given Z-score, the formula to find the raw score \( X \) is:
  • \( X = \mu + Z \times \sigma \)
This formula helps translate a Z-score into a concrete data value in the original scale of measurement. In the given example, the Z-scores -0.934 and 0.934 are converted to raw scores to find the specific limits within the 65% portion of the distribution.

With mean 70 and standard deviation 12, the calculations yield raw scores of approximately 58.792 and 81.208, helping identify where the central 65% of the data falls.

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Most popular questions from this chapter

True/false: Abraham de Moivre, a consultant to gamblers, discovered the normal distribution when trying to approximate the binomial distribution to make his computations easier.

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of \(71 \mathrm{mph}\) and a standard deviation of \(8 \mathrm{mph}\). a. The current speed limit is \(65 \mathrm{mph}\). What is the proportion of vehicles less than or equal to the speed limit? b. What proportion of the vehicles would be going less than \(50 \mathrm{mph}\) ? c. A new speed limit will be initiated such that approximately \(10 \%\) of vehicles will be over the speed limit. What is the new speed limit based on this criterion? d. In what way do you think the actual distribution of speeds differs from a normal distribution?

An automobile manufacturer introduces a new model that averages 27 miles per gallon in the city. A person who plans to purchase one of these new cars wrote the manufacturer for the details of the tests, and found out that the standard deviation is 3 miles per gallon. Assume that in-city mileage is approximately normally distributed. a. What is the probability that the person will purchase a car that averages less than 20 miles per gallon for in-city driving? b. What is the probability that the person will purchase a car that averages between 25 and 29 miles per gallon for in-city driving?

What are the mean and standard deviation of the standard normal distribution? (b) What would be the mean and standard deviation of a distribution created by multiplying the standard normal distribution by 8 and then adding \(75 ?\)

A group of students at a school takes a history test. The distribution is normal with a mean of \(25,\) and a standard deviation of \(4 .\) (a) Everyone who scores in the top \(30 \%\) of the distribution gets a certificate. What is the lowest score someone can get and still earn a certificate? (b) The top \(5 \%\) of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state?

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