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The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly. Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

Short Answer

Expert verified
77.45% of sums are between the specified bounds.

Step by step solution

01

Understand the Problem

We are asked to find the percentage of sums of a sample of 40 drawn from a cholesterol test distribution, which is normally distributed. The percentage is to be determined between two points: 1.5 standard deviations below the mean of these sums and 1 standard deviation above this mean.
02

Calculate the Mean of the Sums

For a normal distribution with a given mean \( \mu = 180 \), the mean of the sums \( \mu_{sum} \) for a sample size \( n = 40 \) is calculated as:\[ \mu_{sum} = n \times \mu \]\[ \mu_{sum} = 40 \times 180 = 7200 \]
03

Calculate the Standard Deviation of the Sums

The standard deviation of the sums \( \sigma_{sum} \) is given by:\[ \sigma_{sum} = \sigma \times \sqrt{n} \]where \( \sigma = 20 \) is the standard deviation of the individual cholesterol test value.\[ \sigma_{sum} = 20 \times \sqrt{40} \approx 126.49 \]
04

Define the Range for the Sums

We need to find the range of values for the sums that are between 1.5 standard deviations below the mean and 1 standard deviation above. This is calculated as:- Lower bound: \( \mu_{sum} - 1.5 \times \sigma_{sum} \)- Upper bound: \( \mu_{sum} + 1 \times \sigma_{sum} \) Calculating these bounds:- Lower Bound: \( 7200 - 1.5 \times 126.49 = 7009.27 \)- Upper Bound: \( 7200 + 1 \times 126.49 = 7326.49 \)
05

Use the Standard Normal Distribution

Convert these sums to standard normal \( Z \)-scores using:\[ Z = \frac{X - \mu_{sum}}{\sigma_{sum}} \]For the lower bound \( 7009.27 \):\[ Z_{lower} = \frac{7009.27 - 7200}{126.49} \approx -1.5 \]For the upper bound \( 7326.49 \):\[ Z_{upper} = \frac{7326.49 - 7200}{126.49} \approx 1 \]
06

Find the Percentage Between the Z-Scores

Consult a standard normal distribution table or use a calculator to find the probabilities:- Probability for \( Z_{lower} = -1.5 \) is approximately 0.0668- Probability for \( Z_{upper} = 1 \) is approximately 0.8413The percentage of sums between these Z-scores is:\[ 0.8413 - 0.0668 = 0.7745 \] or 77.45%.

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Key Concepts

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

Normal Distribution
The Normal Distribution is a fundamental concept in statistics and is often referred to as the "bell curve" due to its shape. It represents a continuous probability distribution where most observations cluster around the mean, and fewer observations are found as you move further away. This symmetric pattern shows that the distribution is well-centered and most of the variations are around the average value.

Key characteristics of a normal distribution include:
  • The mean, median, and mode of the distribution are equal.
  • It's symmetric around the mean.
  • The area under the curve represents the total probability, summing up to 1.
  • About 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
Understanding the normal distribution helps in determining probabilities and making inferences about population data. In the example of the cholesterol test, the distribution is assumed to be normal, allowing us to compute probabilities using the curve.
Z-score Calculation
Z-score calculation is an essential tool in statistics, particularly when analyzing normal distributions. A Z-score tells us how many standard deviations an element is from the mean, transforming any normal distribution to the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

The Z-score formula is:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
  • \(X\) is the value for which you are calculating the Z-score.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation of the distribution.
Z-scores allow comparisons between different data points and calculations of probabilities using the standard normal distribution table. In the cholesterol example, converting the sums of test results into Z-scores made it possible to determine the percentage of sums within a range.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that values tend to be close to the mean, while a high standard deviation indicates a wide range of values.

The formula for standard deviation, usually denoted as \(\sigma\), is:
\[ \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(X_i - \mu)^2} \]
Where:
  • \(X_i\) represents each value in the data set.
  • \(\mu\) is the mean of the data set.
  • \(N\) is the number of values in the data set.
In the context of the cholesterol results, knowing the standard deviation (20) allows for understanding how much the test results typically deviate from the mean (180). For the sums of a sample size, it’s calculated by adjusting for the sample size via \(\sigma_{sum} = \sigma \times \sqrt{n}\), which informs about how much the sums differ from the expected cumulative mean.
Sample Mean
The Sample Mean is a key concept in statistics, particularly when dealing with data from a subset of a larger population. It is the average of all observations in a sample and serves as an estimate of the population mean when full data collection is impractical.

To calculate the sample mean (\(\bar{x}\)), use the formula:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} X_i \]
Where:
  • \(X_i\) are the values in the sample.
  • \(n\) is the sample size.
In exercises like the cholesterol test, the sample mean helps calculate the mean of sums for a sample size, thus furthering statistical analyses and interpretations. Here, the sample mean was used to help determine the expected sum of cholesterol levels over 40 tests \(\mu_{sum} = n \times \mu\), making it possible to identify standard deviations’ bounds and calculate probabilities of certain outcomes.

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Most popular questions from this chapter

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers. a. In words, \(X=\) _____ b. In words, \(X=\) _____ c. \(\quad \overline{X} \sim\) _____(_____,_____) d. Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. e. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. a. If \(X=\) average distance in feet for 49 fly balls, then \(X \sim\) _____(_____,_____) b. What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for \(\overline{X} .\) Shade the region corresponding to the probability. Find the probability. c. Find the \(80^{\text { th }}\) percentile of the distribution of the average of 49 fly balls.

The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited. The \(90^{\text { th }}\) percentile sample average wait time (in minutes) for a sample of 100 riders is: a. 315.0 b. 40.3 c. 38.5 d. 65.2

Richard’s Furniture Company delivers furniture from 10 A.M. to 2 P.M. continuously and uniformly. We are interested in how long (in hours) past the 10 A.M. start time that individuals wait for their delivery. The average wait time is: a. one hour. b. two hours. c. two and a half hours. d. four hours.

A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. Find the probability that the mean actual weight for the 100 weights is greater than 25.2.

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