/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 The proportion of observations f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The proportion of observations from a standard Normal distribution with values less than 1.15 is \(\begin{array}{ll}{\text { (a) } 0.1251 .} & {\text { (c) } 0.8749 .} & {\text { (e) none of these. }} \\ {\text { (b) } 0.8531 .} & {\text { (d) } 0.8944}\end{array}\)

Short Answer

Expert verified
Option (c) 0.8749.

Step by step solution

01

Understand the Problem

We are asked to find the proportion of observations in a standard normal distribution that are less than 1.15. This involves looking up a Z-score in the standard normal distribution table.
02

Review Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. We use Z-scores to find probabilities associated with specific values.
03

Identify the Z-score

The Z-score required is 1.15, which corresponds to the standard score in the standard normal distribution for which we need to find the probability.
04

Use Z-table Lookup

Look up the Z-score of 1.15 in the standard normal distribution table. This table gives the cumulative probability of values less than a given Z-score.
05

Find the Corresponding Probability

From the Z-table, the cumulative probability that Z is less than 1.15 is approximately 0.8749.
06

Choose the Correct Answer

Match the probability from the Z-table (0.8749) with the given options. The choice that matches is option (c) 0.8749.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-score
A Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values within a dataset.
The concept of a Z-score is particularly useful when dealing with a standard normal distribution, which is a special type of normal distribution.
  • The Z-score itself is calculated using the formula:
\[ Z = \frac{(X - \mu)}{\sigma} \]
  • Where \(X\) is the value in question, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation of the distribution.
Once you have a Z-score, you can look it up on a Z-table (or standard normal distribution table) to find the cumulative probability of the score, which tells us the probability of obtaining a result less than the Z-score in the standard normal distribution.
Understanding Z-scores helps in determining how far away a particular observation is from the average (mean). This makes it easier to gauge the likelihood of observing certain values in a dataset.
cumulative probability
Cumulative probability relates to the likelihood that a variable will take a value less than or equal to a specific number.
In the context of the standard normal distribution and Z-scores, cumulative probability gives us crucial information about the distribution of values. To better grasp this, consider the following points:
  • Cumulative probability is determined by integrating the probability density function up to a given point, providing the total probability of observations below that point.
  • Using a standard normal distribution table, cumulative probability can be found for specific Z-scores.
  • The value obtained tells us how common or rare a particular value is within a standard normal distribution.
If a Z-score is 1.15 and the cumulative probability is 0.8749, as the exercise illustrates, it tells us that approximately 87.49% of values lie below 1.15 in the standard normal distribution.
This concept is key when dealing with distributions as it helps to understand the spread and likelihood of observations being below a certain threshold.
standard deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In a normal distribution, it indicates how much individual values differ from the mean of the distribution.
Understanding standard deviation can illuminate insights about data variability.Key elements of standard deviation include:
  • The formula for standard deviation in a sample is:
\[\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2} \]
  • Where \(N\) is the number of observations, \(X_i\) are the individual values, and \(\mu\) is the mean of these values.
  • In a standard normal distribution, the standard deviation is always 1.
A smaller standard deviation indicates that the data points are close to the mean, while a larger one suggests a wider spread.
Understanding standard deviation is vital for interpreting the scale of variability in the dataset and helps in comparing different datasets.
mean
The mean, commonly known as the average, is a critical concept in statistics, representing the central value of a set of numbers.
In terms of distribution, the mean provides a central point, around which other values are spread out.For better clarity, here are the fundamental points about mean:
  • It is calculated by adding up all the values in a dataset and dividing by the number of values, expressed as:
\[ \mu = \frac{\sum_{i=1}^{N} X_i}{N} \]
  • In a standard normal distribution, the mean is always 0, which helps in simplifying the calculation of Z-scores and standard deviation.
The mean provides a quick snapshot of the data and serves as a baseline comparison for individual observations.
Understanding the mean is a foundational element in statistics and helps in dissecting more complex statistical phenomena, allowing students to build more profound insights into the nature of the data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Scores on the ACT test for the 2007 high school graduating class had mean 21.2 and standard deviation 5.0. In all, 1,300,599 students in this class took the test. Of these, 149,164 had scores higher than 27 and another 50,310 had scores exactly 27. ACT scores are always whole numbers. The exactly Normal N(21.2, 5.0) distribution can include any value, not just whole numbers. What’s more, there is no area exactly above 27 under the smooth Normal curve. So ACT scores can be only approximately Normal. To illustrate this fact, find (a) the percent of 2007 ACT scores greater than 27. (b) the percent of 2007 ACT scores greater than or equal to 27. (c) the percent of observations from the N(21.2, 5.0) distribution that are greater than 27. (The percentgreater than or equal to 27 is the same, because there is no area exactly over 27.)

Length of pregnancies The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. For each part, follow the four-step process. (a) At what percentile is a pregnancy that lasts 240 days (that’s about 8 months)? (b) What percent of pregnancies last between 240 and 270 days (roughly between 8 months and 9 months)? (c) How long do the longest 20% of pregnancies last?

The distribution of weights of 9 -ounce bags of a particular brand of potato chips is approximately Normal with mean \(\mu=9.12\) ounces and standard deviation \(\sigma=0.05\) ounce. Draw an accurate sketch of the distribution of potato chip bag weights. Be sure to label the mean, as well as the points one, two, and three standard deviations away from the mean on the horizontal axis.

Outliers The percent of the observations that are classified as outliers by the \(1.5 \times I Q R\) rule is the same in any Normal distribution. What is this percent? Show your method clearly.

Cool pool? Coach Ferguson uses a thermometer to measure the temperature (in degrees Celsius) at 20 different locations in the school swimming pool. An analysis of the data yields a mean of \(25^{\circ} \mathrm{C}\) and a standard deviation of \(2^{\circ} \mathrm{C}\) . Find the mean and standard deviation of the temperature readings in degrees Fahrenheit (recall that \(^{\circ} \mathrm{F}=(9 / 5)^{\circ} \mathrm{C}+32 )\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.