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91Ó°ÊÓ

\(X \sim N(3,5)\) \(\sigma=\)_____.

Short Answer

Expert verified
\(\sigma = \sqrt{5}\)

Step by step solution

01

Understand the Notation

The notation \(X \sim N(\mu, \sigma^2)\) means that \(X\) is a normally distributed random variable with mean \(\mu\) and variance \(\sigma^2\).
02

Extract Information

From the exercise, we know \(X \sim N(3,5)\), which tells us that \(\mu = 3\) and \(\sigma^2 = 5\).
03

Identify the Standard Deviation

The variance \(\sigma^2\) is the square of the standard deviation. Therefore, to find \(\sigma\), we need to take the square root of \(\sigma^2\).
04

Calculate the Standard Deviation

Compute \(\sigma = \sqrt{\sigma^2} = \sqrt{5}\).

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Key Concepts

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

Mean
The term "mean" refers to the average value of a set of numbers. In the context of a normal distribution, the mean, denoted as \(\mu\), is a measure that represents the central tendency of the distribution. In simpler terms, it indicates where the bulk of the numbers lie, or the center of the curve of the distribution.
To calculate the mean of a data set, sum all the values and then divide by the total number of values. Mathematically, if you have a dataset \(X = \{x_1, x_2, \dots, x_n\}\), the mean \(\mu\) is given by: \[ \mu = \frac{x_1 + x_2 + \cdots + x_n}{n} \] In a normal distribution, the mean not only helps describe the distribution's central point but also acts as an axis of symmetry. This means that the distribution curve is ideally the same on both sides of the mean. If we say \(X \sim N(3, 5)\), this indicates that the average value of \(X\) is 3, placing the center of our curve at 3. This is critical for understanding and predicting outcomes based on the distribution.
Variance
Variance is a measure of how spread out the numbers in a data set are. For a normal distribution, it is denoted by \(\sigma^2\). Variance tells us how much the values in the distribution deviate from the mean. A low variance indicates that the values are clustered close to the mean, whereas a high variance indicates that the values are spread out over a wider range.
To calculate variance, follow these steps:
  • Find the mean \(\mu\).
  • Subtract the mean from each value to find the deviation of each value from the mean.
  • Square each deviation.
  • Find the average of these squared deviations.This results in the variance:
\[ \sigma^2 = \frac{(x_1 - \mu)^2 + (x_2 - \mu)^2 + \cdots + (x_n - \mu)^2}{n} \] In our context, \(X \sim N(3, 5)\) gives us a variance of 5. This means that the data points are spread around the mean with a certain level of dispersion. In practical terms, this tells us about the amount of uncertainty or variability in the data.
Standard Deviation
Standard deviation is another way to measure the spread of a distribution. It is the square root of variance and is denoted with the symbol \(\sigma\). Because the standard deviation is in the same unit as the data set, it provides a clearer sense of the spread of values.

Calculating the standard deviation involves these steps:
  • Take the variance \(\sigma^2\).
  • Calculate its square root to find \(\sigma\).
In formula terms:\[\sigma = \sqrt{\sigma^2}\]Thus, for our exercise, we have \(X \sim N(3, 5)\) with \(\sigma^2 = 5\). Therefore, the standard deviation \(\sigma\) is:\[\sigma = \sqrt{5} \]This numerical value tells us the average distance of each data point from the mean. Standard deviation is especially useful when comparing the spread between different data sets, enabling a deeper understanding of the variability present.

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Most popular questions from this chapter

Use the following information to answer the next four exercises: X ~ N(54, 8) Find the probability that \(x>56\)

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. 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. a. If X = distance in feet for a fly ball, then X ~ _____(_____,_____) b. If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement.

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 heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, ? = 79 inches and a standard deviation, ? = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences. a. 77 inches b. 85 inches c. If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer.

We flip a coin 100 times (n = 100) and note that it only comes up heads 20% (p = 0.20) of the time. The mean and standard deviation for the number of times the coin lands on heads is ? = 20 and ? = 4 (verify the mean and standard deviation). Solve the following: a. There is about a 68% chance that the number of heads will be somewhere between ___ and ___. b. There is about a ____chance that the number of heads will be somewhere between 12 and 28. c. There is about a ____ chance that the number of heads will be somewhere between eight and 32.

In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. a. State the approximate distribution of X. b. Is 1,956.8 a population mean or a sample mean? How do you know? c. Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement. d. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton. e. Find the third quartile for votes for President Clinton.

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