/*! 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 14 Find the indicated probabilities... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the indicated probabilities. $$ \mu=100, \sigma=15, \text { find } P(70 \leq X \leq 80) $$

Short Answer

Expert verified
The probability that a data point falls within the interval \(70 \leq X \leq 80\) for a mean of \(\mu = 100\) and a standard deviation of \(\sigma = 15\) is approximately \(0.0690\) or \(6.90\%\).

Step by step solution

01

Computing z-scores for the given values

To compute the z-scores for both the lower limit (\(x_1 = 70\)) and upper limit (\(x_2 = 80\)) of the interval, we use the z-score formula: $$ z = \frac{x - \mu}{\sigma} $$ where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For \(x_1 = 70\): $$z_1 = \frac{70 - 100}{15} = -2$$ For \(x_2 = 80\): $$z_2 = \frac{80 - 100}{15} = -\frac{4}{3}$$
02

Using the standard normal distribution table

Now we will refer to the standard normal distribution table (or a calculator with similar functionality) to find the corresponding probabilities for the calculated z-scores (\(z_1\) and \(z_2\)). Using a standard normal distribution table, we find: $$ P(Z \leq z_1) = P(Z \leq -2) = 0.0228 $$ and $$ P(Z \leq z_2) = P(Z \leq -\frac{4}{3}) = 0.0918 $$
03

Calculate the probability for the interval

To find the probability that a data point lies within the interval \(70 \leq X \leq 80\), we need to find the probability between our z-scores, \(-2 \leq Z \leq -\frac{4}{3}\): $$ P(70 \leq X \leq 80) = P(-2 \leq Z \leq -\frac{4}{3}) = P(Z \leq -\frac{4}{3}) - P(Z \leq -2) $$ Substitute the probabilities from the table or calculator: $$ P(70 \leq X \leq 80) = 0.0918 - 0.0228 = 0.0690 $$ So, the probability that a data point falls within the interval \(70 \leq X \leq 80\) is \(0.0690\) or \(6.90\%\).

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
The z-score is a statistical measure that tells us how many standard deviations a data point is from the mean. It is a way to standardize scores across different datasets or different metrics.
  • To calculate a z-score, you need three pieces of information: the data point (\( x \)), the mean (\( \mu \)), and the standard deviation (\( \sigma \)).
  • The formula is: \[ z = \frac{x - \mu}{\sigma} \]
A z-score can be positive or negative. A positive z-score means the data point is above the mean, while a negative z-score means it is below the mean. Here, a z-score of -2 implies that the data point is 2 standard deviations below the mean.
Understanding z-scores is important because they allow us to determine the probability of a data point falling within a particular range in a normal distribution. This is crucial when analyzing data to make informed decisions.
standard normal distribution
The standard normal distribution is a special normal distribution that has a mean of 0 and a standard deviation of 1. It is represented by the bell-shaped curve known as the Gaussian distribution.
  • Every normal distribution can be transformed into a standard normal distribution using z-scores.
  • When working with the standard normal distribution, probabilities are easier to find because we can use a standard normal distribution table that gives the likelihood of a z-score being less than a given value.
The transformation to a standard normal distribution allows you to compare different sets of data points by converting them to a common scale. Understanding the properties of the standard normal distribution is essential for calculating probabilities and conducting statistical analyses effectively.
If you're ever unsure about a specific probability, consulting the standard normal distribution table or using statistical software will help you find the probability related to your calculated z-scores.
interval probability
Interval probability refers to the probability that a random variable falls within a specific range or interval. This is often expressed as \( P(a \leq X \leq b) \) where \( a \) and \( b \) are the endpoints of that range.
  • Interval probabilities are evaluated by finding the probabilities associated with the z-scores of these endpoints.
  • After converting the original data points into z-scores, the next step is to use the probabilities derived from the standard normal table to identify the likelihood of a variable being within the specified interval.
The key takeaway is to subtract the smaller cumulative probability from the larger one to find the interval probability. This process requires understanding both z-scores and the standard normal distribution.
In practical terms, this helps in determining the likelihood of occurrences within a specific interval, which is essential in fields like finance, quality control, and other areas utilizing statistical inference.

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

The Acme Insurance Company is launching a drive to generate greater profits, and it decides to insure racetrack drivers against wrecking their cars. The company's research shows that, on average, a racetrack driver races four times a year and has a 1 in 10 chance of wrecking a vehicle, worth an average of \(\$ 100,000\), in every race. The annual premium is \(\$ 5,000\), and Acme automatically drops any driver who is involved in an accident (after paying for a new car), but does not refund the premium. How much profit (or loss) can the company expect to earn from a typical driver in a year? HINT [Use a tree diagram to compute the probabilities of the various outcomes.]

Based on data from Nielsen Research, there is a \(15 \%\) chance that any television that is turned on during the time of the evening newscasts will be tuned to \(\mathrm{ABC}\) 's evening news show. \({ }^{58}\) Your company wishes to advertise on a small local station carrying \(\mathrm{ABC}\) that serves a community with 2,500 households that regularly tune in during this time slot. Find the approximate probability that at least 400 households will be tuned in to the show. HINT [See Example 4.]

Stock Market Gyrations Following is a sample of the day-byday change, rounded to the nearest 100 points, in the Dow Jones Industrial Average during 10 successive business days around the start of the financial crisis in October \(2008:^{37}\) \(400,-500,-200,-700,-100,900,-100,-700,400,-100\) a. Compute the mean and standard deviation of the given sample. (Round your answers to the nearest whole number.) b. Assuming the distribution of day-by-day changes of the Dow during financial crises is symmetric and bell-shaped, then the Dow falls by more than \(\longrightarrow\) points \(16 \%\) of the time. What is the percentage of times in the sample that the Dow actually fell by more than that amount? HINT [See

Popularity Ratings In your bid to be elected class representative, you have your election committee survey five randomly chosen students in your class and ask them to rank you on a scale of \(0-10\). Your rankings are \(3,2,0,9,1\). a. Find the sample mean and standard deviation. (Round your answers to two decimal places.) HINT [See Example 1 and Quick Examples on page 581.] b. Assuming the sample mean and standard deviation are indicative of the class as a whole, in what range does the empirical rule predict that approximately \(68 \%\) of the class

Which would you expect to have the greater variance: the standard normal distribution or the uniform distribution taking values between \(-1\) and \(1 ?\) Explain.

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.