/*! 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 18 In a survey of 12 companies recr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 18

In a survey of 12 companies recruiting for recent college graduates, they reported the following numbers of job applicants per job posting: \(123,123,134,127,115,122,125,101,130,143,110,\) and 122 . a. Find the mean and standard deviation, including units. b. What is the \(Z\) score for the company with 143 job applicants per job posting?

Short Answer

Expert verified
Mean = 122.92 applicants, SD = 10.44 applicants; Z-score for 143 applicants = 1.92.

Step by step solution

01

Calculate the Mean

To find the mean, sum up all the reported job applicants and divide by the total number of companies surveyed. The sum of the job applicants is: \[ 123 + 123 + 134 + 127 + 115 + 122 + 125 + 101 + 130 + 143 + 110 + 122 = 1475 \]The number of companies is 12. Thus, the mean (\( \bar{x} \)) is:\[ \bar{x} = \frac{1475}{12} = 122.92 \]The mean number of job applicants per job posting is 122.92.
02

Calculate the Variance

First, find the squared deviations from the mean for each data point, then the variance by averaging these squared differences.The squared deviations are:\[(123 - 122.92)^2, (123 - 122.92)^2, (134 - 122.92)^2, (127 - 122.92)^2, (115 - 122.92)^2, (122 - 122.92)^2, (125 - 122.92)^2, (101 - 122.92)^2, (130 - 122.92)^2, (143 - 122.92)^2, (110 - 122.92)^2, (122 - 122.92)^2\]The individual squared results and their sum is:\[(0.08)^2 + (0.08)^2 + (11.08)^2 + (4.08)^2 + (-7.92)^2 + (-0.92)^2 + (2.08)^2 + (-21.92)^2 + (7.08)^2 + (20.08)^2 + (-12.92)^2 + (-0.92)^2 \]Calculating these yields:\[0.0064, 0.0064, 122.9264, 16.6464, 62.7264, 0.8464, 4.3264, 480.1264, 50.1264, 403.2064, 166.6564, 0.8464\]Their total is:\[ 1308.9 \]Then the variance (\( s^2 \)) is: \[ s^2 = \frac{1308.9}{12} = 109.075\]
03

Calculate the Standard Deviation

The standard deviation is the square root of the variance. Thus:\[ s = \sqrt{109.075} \approx 10.44 \]The standard deviation of job applicants per job posting is approximately 10.44.
04

Calculate the Z Score for 143 Job Applicants

To find the Z score for a company with 143 job applicants, use the formula:\[ Z = \frac{X - \bar{x}}{s} \]Where \(X = 143\), \(\bar{x} = 122.92\), and \(s = 10.44\). Substituting these values gives:\[ Z = \frac{143 - 122.92}{10.44} \approx 1.92 \]The Z score for 143 job applicants is approximately 1.92.

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 91影视!

Key Concepts

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

Understanding Mean Calculation
The mean, also known as the average, is a measure of central tendency that helps us understand the overall trend in a set of numbers. To calculate the mean, simply add up all the numbers and then divide by how many numbers there are.

In this exercise, we are given the number of job applicants per job posting for 12 companies. We add up all the applicants:
  • 123 + 123 + 134 + 127 + 115 + 122 + 125 + 101 + 130 + 143 + 110 + 122 = 1475
Then we divide this sum by the number of companies, which is 12, to find the mean:
  • \[ \bar{x} = \frac{1475}{12} = 122.92 \]
This means that, on average, there are about 122.92 job applicants per job posting.
Exploring Standard Deviation
The standard deviation is a measure that tells us how spread out the numbers in a data set are. It helps us understand how much variation exists from the mean.
To find it, we calculate the variance first and then take the square root.
  • Variance measures the average of the squared differences from the mean.
  • Standard deviation is the square root of the variance, providing a more interpretable number that shares the same unit as the data.
In our exercise, after calculating the variance to be 109.075, the standard deviation is:
  • \[ s = \sqrt{109.075} \approx 10.44 \]
Thus, a standard deviation of approximately 10.44 indicates that the typical difference between each company鈥檚 number of job applicants and the mean is about 10.44 applicants.
Interpreting Z Score
A Z score tells us how many standard deviations away a data point is from the mean. It is a great way to understand the position of a single data point within a data set.
To compute the Z score, use the formula:
  • \[ Z = \frac{X - \bar{x}}{s} \]
where:
  • \(X\) is the data point
  • \(\bar{x}\) is the mean
  • \(s\) is the standard deviation
For a company with 143 job applicants, the Z score calculates as:
  • \[ Z = \frac{143 - 122.92}{10.44} \approx 1.92 \]
This Z score of approximately 1.92 signifies that 143 applicants are 1.92 standard deviations above the mean number of job applicants per posting.
Diving into Variance Calculation
Variance quantifies how much the numbers in a data set differ from the mean. Larger variance indicates numbers are spread out, while smaller variance suggests they are close to the mean.

Steps to calculate variance include:
  • Find the squared difference between each data point and the mean.
  • Average these squared differences.
In the given problem, calculating the squared deviations, we get:
  • Sum of squared deviations \( = 0.0064 + 0.0064 + 122.9264 + 16.6464 + 62.7264 + 0.8464 + 4.3264 + 480.1264 + 50.1264 + 403.2064 + 166.6564 + 0.8464 = 1308.9 \)
Then, we find the variance:
  • \[ s^2 = \frac{1308.9}{12} = 109.075 \]
Therefore, the variance of 109.075 showcases the extent of spread among job applicants for the dataset.

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

An experiment compared the ability of three groups of participants to remember briefly-presented chess positions. The data are shown below. The numbers represent the average number of pieces correctly remembered from three chess positions. a. Find the standard deviation of each group. b. Calculate the 5 -number summary for each group. c. Calculate the range and IQR for each group. d. Create side-by-side boxplots and compare and contrast the two groups. $$ \begin{array}{|c|c|c|} \hline \text { Non-players } & \text { Beginners } & \text { Tournament Players } \\ \hline 22.1 & 32.5 & 40.1 \\ \hline 22.3 & 37.1 & 45.6 \\ \hline 26.2 & 39.1 & 51.2 \\ \hline 29.6 & 40.5 & 56.4 \\ \hline 31.7 & 45.5 & 58.1 \\ \hline 33.5 & 51.3 & 71.1 \\ \hline 38.9 & 52.6 & 74.9 \\ \hline 39.7 & 55.7 & 75.9 \\ \hline 39.7 & 55.7 & 75.9 \\ \hline 43.2 & 55.9 & 80.3 \\ \hline 43.2 & 57.7 & 85.3 \\ \hline \end{array} $$

What types of graphs are used for quantitative data?

To determine the average length of trout in a lake, researchers catch 20 fish and measure them. Describe the population and sample of this study.

A random sample of 45 people who carry a purse found that they had an average of $$\$ 2.35$$ in change in the bottom of their purse. The margin of error was $$\$ 0.15 .$$ Calculate the \(95 \%\) confidence interval and interpret the results.

Use technology to answer these questions. Suppose a Normal distribution has a mean of 26.1 grams and a standard deviation of 6.5 grams. a. Draw and label the Normal distribution graph. b. What percentage of the data values fall above 32.6 grams? c. What percentage of data is below 15 grams or greater than 36.7 grams? d. What percentage of the data is less than or equal to 20.8 grams?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation
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.