/*! 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 58 The mean number of text messages... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The mean number of text messages sent per month by customers of a cell phone service provider is 1,650 , and the standard deviation is \(750 .\) Find the \(z\) -score associated with each of the following numbers of text messages sent. a. 0 b. \(\quad 10,000\) c. \(\quad 4,500\) d. \(\quad 300\)

Short Answer

Expert verified
The z-scores for 0, 10,000, 4,500 and 300 text messages sent are -2.2, 11.13, 3.8 and -1.8 respectively.

Step by step solution

01

Understand the z-score formula

The z-score formula is given as \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value for which we want to find the z-score, \(\mu\) is the mean and \(\sigma\) is the standard deviation.
02

Calculate z-score for 0 messages

Substitute \(x = 0\), \(\mu = 1650\) and \(\sigma = 750\) in the z-score formula. The z-score, \(z = \frac{0 - 1650}{750} = -2.2\).
03

Calculate z-score for 10,000 messages

Substitute \(x = 10000\), \(\mu = 1650\) and \(\sigma = 750\) in the z-score formula. The z-score, \(z = \frac{10000 - 1650}{750} = 11.13\).
04

Calculate z-score for 4,500 messages

Substitute \(x = 4500\), \(\mu = 1650\) and \(\sigma = 750\) in the z-score formula. The z-score, \(z = \frac{4500 - 1650}{750} = 3.8\).
05

Calculate z-score for 300 messages

Substitute \(x = 300\), \(\mu = 1650\) and \(\sigma = 750\) in the z-score formula. The z-score, \(z = \frac{300 - 1650}{750} = -1.8\).

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.

Understanding the Mean
The mean, often referred to as the average, is an essential concept in statistics. It helps us understand the central value of a data set. To calculate the mean, you simply add up all the values and then divide by the number of values. This gives you an idea of what a "typical" value might be in your data set. In the context of our exercise, the mean number of text messages sent per month is 1,650. Here are some important aspects of the mean:
  • If all values in a data set are equal, the mean will be that common value.
  • The mean is sensitive to extreme values (outliers).
  • It is used as a reference point for calculating the z-score, as seen in our exercise.
Understanding the mean is crucial because it helps in understanding the overall behavior of the data, especially when dealing with larger datasets.
Delving into Standard Deviation
Standard deviation is a powerful measure of how spread out the values in a data set are. It tells us on average how much each value differs from the mean. This can help us understand the variability or consistency within a data set. Calculating standard deviation involves a few steps:
  • Find the difference between each value and the mean.
  • Square each difference to eliminate negative values.
  • Calculate the average of these squared differences.
  • The square root of this average gives the standard deviation.
In our exercise, the standard deviation of text messages sent is 750. A large standard deviation, like in this example, indicates that the number of messages sent varies significantly month to month. Knowing the standard deviation helps us determine how typical or unusual a particular data point is, which is why it's used in calculating the z-score.
Exploring Normal Distribution
Normal distribution is a vital concept in statistics. It's often depicted as a bell-shaped curve that suggests how data points are expected to spread out or cluster around the mean in a data set. Characteristics of a normal distribution include:
  • The curve is symmetric around the mean.
  • The mean, median, and mode of the data set are equal and lie at the center.
  • A large proportion of data points are near the mean, with fewer points further out.
  • About 68% of observations lie within one standard deviation of the mean.
  • About 95% fall within two standard deviations.
  • Nearly all (99.7%) fall within three standard deviations.
In probability and statistics, many natural phenomena are assumed to follow a normal distribution. Understanding this helps interpret how likely we are to observe certain values, such as in our exercise where understanding how text message numbers deviate from the mean is important. The concept of normal distribution is also integral to the calculation of the z-score, transforming individual data points into standardized values, which allow for comparisons.

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 paper "Total Diet Study Statistics on Element Results" (Food and Drug Administration, April 25,2000 ) gave information on sodium content for various types of foods. Twenty-six brands of tomato catsup were analyzed. Data consistent with summary quantities in the paper are Sodium content ( \(\mathrm{mg} / \mathrm{kg}\) ) \(\begin{array}{lrrrrr}12,148 & 10,426 & 10,912 & 9,116 & 13,226 & 11,663 \\ 11,781 & 10,680 & 8,457 & 10,788 & 12,605 & 10,591 \\\ 11,040 & 10,815 & 12,962 & 11,644 & 10,047 & \\ 10,478 & 10,108 & 12,353 & 11,778 & 11,092 & \\ 11,673 & 8,758 & 11,145 & 11,495 & & \end{array}\) Calculate and interpret the values of the quartiles and the interquartile range. (Hint: See Example 3.9 )

The San Luis ObispoTelegram-Tribune(October1,1994) reported the following monthly salaries for supervisors from six different counties: \(\$ 5,354\) (Kern), \(\$ 5,166\) (Monterey), \(\$ 4,443\) (Santa Cruz), \(\$ 4,129\) (Santa Barbara), \(\$ 2,500\) (Placer), and \$2,220 (Merced). San Luis Obispo County supervisors are supposed to be paid the average of the two counties in the middle of this salary range. Which measure of center determines this salary, and what is its value? Find the value of the other measure of center featured in this chapter. Why is it not as favorable to the San Luis Obispo County supervisors (although it might appeal to taxpayers)?

The report "Earnings by Education Level and Gender" (U.S. Census Bureau, 2009) gave the following percentiles for annual earnings of full-time female workers age 25 and older with an associate degree: $$ \begin{array}{l} \text { 25th percentile }=\$ 26,800 \text { 50th percentile }=\$ 36,800 \\ \text { 75th percentile }=\$ 51,100 \end{array} $$ a. For each of these percentiles, write a sentence interpreting the value of the percentile. b. The report also gave percentiles for men age 25 and older with an associate degree: $$ \begin{array}{l} \text { 25th percentile }=\$ 35,700 \text { 50th percentile }=\$ 50,100 \\ \text { 75th percentile }=\$ 68,000 \end{array} $$ Write a few sentences commenting on how these values compare to the percentiles for women.

Data on tipping percent for 20 restaurant tables, consistent with summary statistics given in the paper "Beauty and the Labor Market: Evidence from Restaurant Servers"(unpublished manuscript by Matt Parrett, 2007), are: \(\begin{array}{rrrrrrr}0.0 & 5.0 & 45.0 & 32.8 & 13.9 & 10.4 & 55.2 \\ 50.0 & 10.0 & 14.6 & 38.4 & 23.0 & 27.9 & 27.9 \\ 105.0 & 19.0 & 10.0 & 32.1 & 11.1 & 15.0 & \end{array}\) a. Calculate the mean and standard deviation for this data set. b. Delete the observation of 105.0 and recalculate the mean and standard deviation. How do these values compare to the values from Part (a)? What does this suggest about using the mean and standard deviation as measures of center and spread for a data set with outliers?

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the following data on time (in seconds) to complete the escape ("Oxygen Consumption and Ventilation During Escape from an Offshore Platform," Ergonomics [1997]: \(281-292\) ): \(\begin{array}{lllllllll}389 & 356 & 359 & 363 & 375 & 424 & 325 & 394 & 402 \\\ 373 & 373 & 370 & 364 & 366 & 364 & 325 & 339 & 393 \\ 392 & 369 & 374 & 359 & 356 & 403 & 334 & 397 & \end{array}\) a. Construct a dotplot of the data. Will the mean or the median be larger for this data set? b. Calculate the values of the mean and median. c. By how much could the largest time be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the median?

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.