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In a normal distribution with \(\mu=0\) and \(\sigma=4,\) find the \(x\) -value that corresponds to the a) 50 th percentile b) 84 th percentile

Short Answer

Expert verified
50th percentile: \(x = 0\); 84th percentile: \(x = 4\).

Step by step solution

01

Identifying the Question

We need to find the value for the normal variable, \(x\), which corresponds to the provided percentiles (50th and 84th) given a normal distribution with mean \(\mu=0\) and standard deviation \(\sigma=4\).
02

Understanding the 50th Percentile

The 50th percentile in a normal distribution is the median, which equals the mean \(\mu\).
03

Calculating the 50th Percentile

Since \(\mu=0\), the \(x\)-value for the 50th percentile is \(x=0\).
04

Understanding the 84th Percentile

To find the \(x\)-value for the 84th percentile, we first determine the corresponding \(z\)-score using the standard normal distribution table. The 84th percentile approximately corresponds to a \(z\)-score of 1.
05

Calculating the 84th Percentile

Use the \(z\)-score formula \(x = \mu + z\sigma\). Here, \(z=1\), \(\mu=0\), and \(\sigma=4\). Substitute to find the \(x\)-value: \[x = 0 + 1 \times 4 = 4\]

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Key Concepts

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

Percentiles
Percentiles are a way to understand the position of a value within a distribution. Imagine you have a hundred people lined up in order of their height. The 50th percentile would be the height of the person right in the middle. In any dataset, a percentile indicates that a certain percentage of the data points fall below that value. In the context of a normal distribution, percentiles help you locate where a particular value lies compared to the overall data set. For instance, the 50th percentile, also known as the median, indicates that half of the values are below, and half are above it.
  • The 50th percentile is the same as the median in a symmetric distribution.
  • Understanding percentiles can help determine cutoff points for tests and scores.
  • High percentiles (e.g., 90th) mean that a value is higher than most other values.
To find the specific value at a given percentile in a normal distribution, you often use statistical tables or software tools. This allows you to determine where your particular data point stands in relation to the whole distribution.
Z-Score
A Z-score tells you how far a specific value is from the mean of a distribution, measured in standard deviations. It's a useful way to determine how unusual or typical a data point is compared to other points in the distribution. If you know the mean and standard deviation, you can calculate the Z-score. The formula for the Z-score is:\[ z = \frac{x - \mu}{\sigma} \]Where:
  • \(z\) is the Z-score,
  • \(x\) is the value being evaluated,
  • \(\mu\) is the mean of the distribution, and
  • \(\sigma\) is the standard deviation.
A Z-score of 0 means the value is exactly at the mean, while a positive Z-score indicates it is above the mean and a negative Z-score indicates it is below. The Z-score helps in normalizing the distribution, making it easier to calculate probabilities and compare different data sets. In practical terms, Z-scores are often used to look up standard tables to find percentiles, as seen in the calculation of the 84th percentile which corresponds to a Z-score of 1.
Standard Deviation
Standard deviation is a measure of how spread out numbers in a data set are. It is crucial for understanding the distribution of data around the mean. A small standard deviation means that data points are close to the mean, whereas a large standard deviation indicates that data points are spread out over a wider range of values. To calculate standard deviation, you typically follow these steps:
  • Calculate the mean (average) of the data set.
  • Subtract the mean from each data point and square the result.
  • Find the average of these squared differences.
  • Take the square root of the average.
This calculated value gives you the standard deviation, denoted by \(\sigma\) in the context of normal distributions. Standard deviation plays a key role in determining the shape of a normal distribution. For example, in a normal distribution with a mean of 0 and a standard deviation of 4, as noted in the exercise, most values will lie within 4 units of 0. This framework allows for objective assessments of how data is distributed around the average.

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Most popular questions from this chapter

Suppose 30 sparrows are released into a region where they have no natural predators. The growth of the region's sparrow population can be modeled by the uninhibited growth model \(d P / d t=k P\) where \(P(t)\) is the population of sparrows \(t\) years after their initial release. a) When the sparrow population is estimated at \(12,500,\) its rate of growth is about 1325 sparrows per year. Use this information to find \(k,\) and then find the particular solution of the differential equation. b) Find the number of sparrows after 70 yr. c) Without using a calculator, find \(P^{\prime}(70) / P(70)\).

(a) write a differential equation that models the situation, and (b) find the general solution. If an initial condition is given, find the particular solution. Recall that when \(y\) is directly proportional to \(x,\) we have \(y=k x\), and when \(y\) is inversely proportional to \(x,\) we have \(y=k / x,\) where \(k\) is the constant of proportionality. In these exercises, let \(k=1\). The rate of change of \(y\) with respect to \(x\) is inversely proportional to the cube of \(y\).

For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{1}{4}, \quad[3,7] $$

Explain why a normal distribution may not apply if you are analyzing the distribution of weights of students in a classroom.

Before \(1859,\) rabbits did not exist in Australia. That year, a settler released 24 rabbits into the wild. Without natural predators, the growth of the Australian rabbit population can be modeled by the uninhibited growth model \(d P / d t=k P,\) where \(P(t)\) is the population of rabbits \(t\) years after \(1859 .\) (Source: www dpi.vic.gov.au/agriculture.) a) When the rabbit population was estimated to be \(8900,\) its rate of growth was about 2630 rabbits per year. Use this information to find \(k,\) and then find the particular solution of the differential equation. b) Find the rabbit population in \(1900(t=41)\) and the rate at which it was increasing in that year. c) Without using a calculator, find \(P^{\prime}(41) / P(41)\).

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