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Chapter 6: Problem 20

For a normal distribution, a. Find the \(z\) -score for which a total probability of 0.04 falls more than \(z\) standard deviations (in either direction) from the mean, that is, below \(\mu-z \sigma\) or above \(\mu+z \sigma\) b. For this \(z\) -score, explain why the probability of values more than \(z\) standard deviations below the mean is 0.02 . c. Explain why \(\mu+z \sigma\) is the 2 nd percentile.

Short Answer

Expert verified
The required \(z\)-score is 2.05. The probability below \(\mu - z\sigma\) is 0.02. \(\mu + z\sigma\) marks the 2nd percentile.

Step by step solution

01

Understanding the Problem

We're dealing with a normal distribution and need to find a symmetric interval around the mean (\(\mu\)) such that the probability outside this interval is 0.04. We'll determine the \(z\)-score where \(0.04\) of the data lies outside the interval \((\mu - z\sigma, \mu + z\sigma)\).
02

Finding the Total Probability for Two Tails

The problem states that a total probability of \(0.04\) lies outside the range. This means \(0.02\) of the probability is in each tail, below \(\mu - z\sigma\) and above \(\mu + z\sigma\).
03

Using the Standard Normal Distribution Table

From a standard normal distribution table, find the \(z\)-score that corresponds to a cumulative probability of \(0.02\) in the left tail. Using this, we find \(z \approx -2.05\). Since the distribution is symmetric, \(z \approx 2.05\) will have \(0.02\) in the right tail.
04

Part A Solution

The \(z\)-score for which a total probability of 0.04 falls outside is approximately \(2.05\). This means \(\mu - 2.05\sigma\) and \(\mu + 2.05\sigma\) mark the boundaries for \(0.04\) to be outside.
05

Part B Explanation

The \(z\)-score found is \(2.05\), and since the tails are symmetric, \(\mu - 2.05\sigma\) has \(0.02\) probability. This is because we split the total probability outside into \(0.02\) for each tail.
06

Part C Explanation

The value at \(\mu + z \sigma\) corresponds to the cumulative probability of \(0.98\), or the 98th percentile. Conversely, \(\mu - z\sigma\) corresponds to the 2nd percentile because it represents the lower cutoff for \(0.02\) probability.

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Key Concepts

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

Z-score
In the realm of statistics, particularly when discussing the normal distribution, the concept of a z-score is fundamental. A z-score represents the number of standard deviations a data point is from the mean. This allows comparison of data points from different normal distributions.
The calculation of a z-score is straightforward:
  • It is calculated as: \( z = \frac{(X - \mu)}{\sigma} \), where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • A positive z-score indicates the data point is above the mean, while a negative z-score indicates it is below the mean.
Understanding z-scores can help determine how unusual or usual a particular standard deviation is in a distribution, making it a powerful tool for statistical analysis.
Standard Deviations
The concept of standard deviation is crucial to understanding how data is spread around the mean in a set of values. In a normal distribution, data falls within predictable intervals defined by standard deviations.
  • Standard deviation (\(\sigma\)) measures the amount of variation or dispersion in a set of values.
  • A small standard deviation means that the data points tend to be close to the mean.
  • A larger standard deviation indicates that the data points are spread out over a wider range of values.
When examining normal distributions, standard deviations help define the spread and thus allow us to interpret what is normal or abnormal within the dataset. It's also integral to calculating z-scores and understanding the positioning of data points.
Percentile
Percentiles provide insights into how a particular value compares to all other values in a data set. They are particularly useful in understanding how a z-score fits within the context of a normal distribution.
  • A percentile indicates the value below which a given percentage of observations in a group falls.
  • For example, the 2nd percentile is the value below which 2% of the observations may be found.
  • In a normal distribution, percentiles can help identify cutoff points for outliers or extreme values.
In the context of z-scores, the location of the z-score within the distribution can be represented as a percentile, which helps visualize its position relative to the rest of the data.
Cumulative Probability
Cumulative probability is a vital concept in statistics, representing the likelihood that a random variable is less than or equal to a particular value. In normal distributions, cumulative probabilities are used to determine the probability of a z-score.
  • Cumulative probability increases as you move from left to right on a probability curve.
  • It provides the probability that a random variable lies below a particular value.
  • Cumulative distribution functions can show the probability that a variable will fall within a certain range.
When using a z-score table or cumulative distribution function, calculations reveal how likely it is for a value to be below or above a certain point on the curve, forming the basis for many statistical conclusions.

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Most popular questions from this chapter

Ideal number of children Let \(X\) denote the response of a randomly selected person to the question, "What is the ideal number of children for a family to have?" The probability distribution of \(X\) in the United States is approximately as shown in the table, according to the gender of the person asked the question. $$ \begin{array}{lcc} \hline \begin{array}{l} \text { Probability Distribution of } \boldsymbol{X}=\text { Ideal Number } \\\ \text { of Children } \end{array} \\ \hline \boldsymbol{x} & \mathbf{P}(\boldsymbol{x}) \text { Females } & \mathbf{P}(\boldsymbol{x}) \text { Males } \\ \hline 0 & 0.01 & 0.02 \\ 1 & 0.03 & 0.03 \\ 2 & 0.55 & 0.60 \\ 3 & 0.31 & 0.28 \\ 4 & 0.11 & 0.08 \\ \hline \end{array} $$ a. Show that the means are similar, 2.50 for females and 2.39 for males. b. The standard deviation for the females is 0.770 and 0.758 for the males. Explain why a practical implication of the values for the standard deviations is that males hold slightly more consistent views than females about the ideal family size.

Unfair wealth distribution sentiment According to a study published in www.gallup.com in \(2015,63 \%\) of Americans said wealth should be more evenly distributed among a larger percentage of people. For a sample of 10 Americans, let \(X=\) number of respondents who said wealth was unfairly distributed. a. Explain why the conditions are satisfied for \(X\) to have the binomial distribution. b. Identify \(n\) and \(p\) for the binomial distribution. c. Find the probability that two Americans in the sample said wealth should be more evenly distributed.

A World Health Organization study (the MONICA project) of health in various countries reported that in Canada, systolic blood pressure readings have a mean of 121 and a standard deviation of \(16 .\) A reading above 140 is considered high blood pressure. a. What is the \(z\) -score for a blood pressure reading of \(140 ?\) b. If systolic blood pressure in Canada has a normal distribution, what proportion of Canadians suffers from high blood pressure? c. What proportion of Canadians has systolic blood pressures in the range from 100 to \(140 ?\) d. Find the 90 th percentile of blood pressure readings.

Profit and the weather From past experience, a wheat farmer living in Manitoba, Canada, finds that his annual profit (in Canadian dollars) is 80,000 if the summer weather is typical, 50,000 if the weather is unusually dry, and 20,000 if there is a severe storm that destroys much of his crop. Weather bureau records indicate that the probability is 0.70 of typical weather, 0.20 of unusually dry weather, and 0.10 of a severe storm. Let \(X\) denote the farmer's profit next year. a. Construct a table with the probability distribution of \(X\) b. What is the probability that the profit is 50,000 or less? c. Find the mean of the probability distribution of \(X\). Interpret. d. Suppose the farmer buys insurance for 3000 that pays him 20,000 in the event of a severe storm that destroys much of the crop and pays nothing otherwise. Find the probability distribution of his profit. Find the mean and summarize the effect of buying this insurance.

For a normal distribution, use Table A, software, or a calculator to find the probability that an observation is a. at least 1 standard deviation above the mean. b. at least 1 standard deviation below the mean. c. within 1 standard deviation of the mean.
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