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Chapter 6: Problem 53

Use the following information to answer the next four exercises: X ~ N(54, 8) Find the \(60^{\text { th }}\) percentile.

Short Answer

Expert verified
The 60th percentile is approximately 56.024.

Step by step solution

01

Understand the Parameters

The exercise provides a normal distribution, with mean \( \mu = 54 \) and standard deviation \( \sigma = 8 \). We need to find the \( 60^{\text{th}} \) percentile of this distribution.
02

Use Z-Score Formula

The formula for the Z-score in a normal distribution is \( Z = \frac{X - \mu}{\sigma} \). To find the \( 60^{\text{th}} \) percentile, we need to know the Z-value corresponding to the \( 0.60 \) percentile.
03

Lookup Z-Score for 60th Percentile

In a standard normal distribution table, a \( 60^{\text{th}} \) percentile corresponds to a Z-score of approximately \( 0.253 \).
04

Calculate the Percentile Value

Using the Z-score formula, solve for \( X \): \[ X = Z \cdot \sigma + \mu \]. Substitute \( Z = 0.253 \), \( \sigma = 8 \), and \( \mu = 54 \) into the equation: \[ X = 0.253 \cdot 8 + 54 = 2.024 + 54 = 56.024 \].
05

Conclusion

The \( 60^{\text{th}} \) percentile of this distribution is approximately \( 56.024 \).

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Key Concepts

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

Percentile Calculation
Understanding percentiles is crucial in statistics. A percentile tells you the value below which a given percentage of observations fall in a data distribution.
The process of calculating a percentile in a normal distribution involves finding the value of the data point where the cumulative distribution function (CDF) equals the desired percentile.

Here's a simplified approach:
  • Identify the desired percentile. In our case, it's the 60th percentile.
  • Convert the percentile to a decimal. For the 60th percentile, this would be 0.60.
  • Use a Z-table or calculator to find the Z-score corresponding to this decimal.
The percentile allows you to understand where a particular data point stands in comparison to the entire dataset; it is a relative measure and does not imply the actual value on its own, but rather its position in the dataset.
Z-score
The Z-score, also known as the standard score, measures how many standard deviations an element is from the mean.

Using the formula:
  • \[ Z = \frac{X - \mu}{\sigma} \]
Z-scores allow comparing different data points from different normal distributions, converting them into a standard normal distribution. This transformation standardizes values across different scales.

For example, suppose you want to find out which test score is better: a score of 85 in one test or 90 in another, where each test has different means and standard deviations. Z-scores make this comparison possible! It's just about seeing how far a data point lies from the average in units of standard deviations.
Standard Deviation
Standard deviation is a fundamental concept in statistics that quantifies the amount of variation or dispersion in a dataset.
It tells us how much individual data points deviate from the mean value, serving as a measure of volatility.

The formula for standard deviation is:
  • \[ \sigma = \sqrt{ \frac{1}{N}\sum_{i=1}^{N} (X_i - \mu)^2 } \]
Where:
  • \( \sigma \) is the standard deviation
  • \( N \) is the number of observations
  • \( X_i \) are the individual data points
  • \( \mu \) is the mean
A larger standard deviation indicates more spread out data. In a normal distribution, most data points lie within one standard deviation of the mean (approximately 68%). If you see a large standard deviation, it's a sign that the data points are more dispersed from the mean.
Mean Value
The mean value, commonly referred to as the "average," is central to many statistical analyses.
It provides a single value that summarizes a dataset by dividing the sum of all values by the number of observations.

Mathematically, it's expressed as:
  • \[ \mu = \frac{1}{N}\sum_{i=1}^{N} X_i \]
Where:
  • \( \mu \) is the mean
  • \( N \) is the number of observations
  • \( X_i \) are the individual data points
The mean is especially useful in normal distributions because it lies at the center of the distribution and acts as a point of symmetry.
It provides a basis for calculating both the standard deviation and Z-scores, making it an essential concept in understanding and working with data.

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Most popular questions from this chapter

A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10% of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68-95-99.7 empirical rule (one standard deviation, two standard deviations and three standard deviations from the mean are being referred to)? Assume a normal distribution for the defective cars in the sample.

A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = _____.

Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Define the random variable X in words. X = _______________.

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. The 90th percentile for recovery times is? a. 8.89 b. 7.07 c. 7.99 d. 4.32

\(X \sim N(-2,1)\) \(\mu=\)_____.
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