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Chapter 11: Problem 141

Find the specified areas for a normal density. (a) The area below 80 on a \(N(75,10)\) distribution (b) The area above 25 on a \(N(20,6)\) distribution (c) The area between 11 and 14 on a \(N(12.2,1.6)\) distribution

Short Answer

Expert verified
(a) The area below 80 on a \(N(75,10)\) distribution is 69.15%. (b) The area above 25 on a \(N(20,6)\) distribution is 20.33%. (c) The area between 11 and 14 on a \(N(12.2,1.6)\) distribution is 64.2%.

Step by step solution

01

Calculate Z-Scores

A Z-Score is how far and in what direction the raw score deviates from the mean, measured in units of standard deviation. For each value given, the Z-Score is calculated as follows: \(Z = \frac{X - \mu}{\sigma}\).\n(a) \(Z = \frac{80 - 75}{10} = 0.5\)\n(b) \(Z = \frac{25 - 20}{6} = 0.83\)\n(c) \(Z1 = \frac{11 - 12.2}{1.6} = -0.75; Z2 = \frac{14 - 12.2}{1.6} = 1.125\)
02

Use Standard Normal Distribution to find Probabilities

Once the Z-scores have been calculated, the standard normal distribution table can be used to find the probabilities.\n(a) The area to the left of 0.5, which is \(P(Z < 0.5)\) is 0.6915.\n(b) The area to the right of 0.83, which is \(P(Z > 0.83)\) is 1 - 0.7967 = 0.2033.\n(c) The area between -0.75 and 1.125, is \(P(-0.75 < Z < 1.125)\) = \(P(Z < 1.125) - P(Z < -0.75)\) = 0.8686 - 0.2266 = 0.642.
03

Convert Probabilities to Percentages

You can turn these probabilities into percentages by multiplying by 100.\n(a) 0.6915 x 100 = 69.15%\n(b) 0.2033 x 100 = 20.33%\n(c) 0.642 x 100 = 64.2%

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Key Concepts

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

Z-Score Calculation
Understanding the concept of Z-Score is pivotal in statistics when dealing with normal distributions. A Z-Score reveals how many standard deviations a data point is from the mean. To calculate a Z-Score, you use the formula:

\( Z = \frac{X - \mu}{\sigma} \),

where \(X\) is a value in the distribution, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For example, if an exam score is 80, the mean score is 75, and the standard deviation is 10, the Z-Score would be \(\frac{80 - 75}{10} = 0.5\). This indicates that the score is 0.5 standard deviations above the mean. Z-Scores are useful because they allow for comparison between different data sets and can help identify outliers.
Standard Deviation
The standard deviation (\(\sigma\)) is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean of the set, while a high standard deviation means that the values are spread out over a wider range.

In the context of a normal distribution, which is symmetric and bell-shaped, about 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% within three. Knowing the standard deviation allows us to calculate Z-Scores and ultimately understand where a particular score lies within the distribution.
Probability Distributions
Probability distributions are mathematical functions that provide the probabilities of occurrence of different possible outcomes. The normal distribution, one of the most common probability distributions, is characterized by its bell shape. Normal distributions are defined by two parameters: the mean (\(\mu\)), which centers the curve, and the standard deviation (\(\sigma\)), which controls the spread.

Every normal distribution can transform into the standard normal distribution, which has a mean of 0 and a standard deviation of 1. We perform this transformation by calculating the Z-Score, thus standardizing different normal distributions and making it possible to use the standard normal distribution table for finding probabilities.
Areas Under the Curve
When we talk about 'areas under the curve' in the context of normal distributions, we are referring to probabilities. The total area under a normal distribution curve represents all possible outcomes and is equal to 1, or 100%. The area under the curve to the left of a Z-Score represents the probability that a variable will fall below that value.

For example, if we want to find the probability of a score being below 80 in a distribution with mean 75 and standard deviation 10, we calculate the Z-Score and then look up this Z-Score in the standard normal distribution table. This table shows us the area to the left of the Z-Score, which is the probability we're seeking. If we want the area above a certain value, we subtract the table value from 1, as probabilities always add up to 100%.

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Most popular questions from this chapter

In Exercises \(\mathrm{P} .78\) to \(\mathrm{P} .81,\) use the probability function given in the table to calculate: (a) The mean of the random variable (b) The standard deviation of the random variable $$ \begin{array}{lccc} \hline x & 10 & 20 & 30 \\ \hline p(x) & 0.7 & 0.2 & 0.1 \\ \hline \end{array} $$

As in Exercise \(\mathrm{P} .35,\) we have a bag of peanut \(\mathrm{M} \& \mathrm{M}\) 's with \(80 \mathrm{M} \& \mathrm{Ms}\) in it, and there are 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is yellow? (b) If we select one at random, what is the probability that it is not brown? (c) If we select one at random, what is the probability that it is blue or green? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are red? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is yellow and the second one is blue?

Random Samples of College Degree Proportions In Exercise P.162, we see that the distribution of sample proportions of US adults with a college degree for random samples of size \(n=500\) is \(N(0.325,0.021) .\) How often will such samples have a proportion, \(\hat{p},\) that is more than \(0.35 ?\)

Find the specified areas for a \(N(0,1)\) density. (a) The area below \(z=0.8\) (b) The area above \(z=1.2\) (c) The area between \(z=-1.75\) and \(z=-1.25\)

State whether the two events (A and B) described are disjoint, independent, and/or complements. (It is possible that the two events fall into more than one of the three categories, or none of them.) South Africa plays Australia for the championship in the Rugby World Cup. At the same time, Poland plays Russia for the World Team Chess Championship. Let \(\mathrm{A}\) be the event that Australia wins their rugby match and \(\mathrm{B}\) be the event that Poland wins their chess match.
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