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In a survey of 106 senior information technology and data security professionals at major U.S. companies regarding their confidence that they had detected all significant security breaches in the past year, the following responses were obtained. $$\begin{array}{lcccc} \hline & \begin{array}{c} \text { Very } \\ \text { confident } \end{array} & \begin{array}{c} \text { Moderately } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not very } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not at all } \\ \text { confident } \end{array} \\ \hline \text { Respondents } & 21 & 56 & 22 & 7 \\ \hline \end{array}$$ What is the probability that a respondent in the survey selected at random a. Had little or no confidence that he or she had detected all significant security breaches in the past year? b. Was very confident that he or she had detected all significant security breaches in the past year?

Short Answer

Expert verified
a. The probability that a respondent in the survey selected at random had little or no confidence that he or she had detected all significant security breaches in the past year is \(\frac{29}{106}\). b. The probability that a respondent in the survey selected at random was very confident that he or she had detected all significant security breaches in the past year is \(\frac{21}{106}\).

Step by step solution

01

Identify the variables

In this scenario, we have the following variables: - Total number of respondents: 106 - Number of respondents with little or no confidence (not very confident + not at all confident): 22 + 7 - Number of respondents who are very confident: 21
02

Calculate the probability for part a

To calculate the probability of a randomly chosen respondent having little or no confidence that they detected all significant security breaches in the past year, we will divide the number of respondents with little or no confidence by the total number of respondents: \(P(\text{Little or No Confidence}) = \frac{\text{Number of Respondents with Little or No Confidence}}{\text{Total Number of Respondents}}\) \(P(\text{Little or No Confidence}) = \frac{22 + 7}{106} = \frac{29}{106}\)
03

Calculate the probability for part b

Similarly, to calculate the probability of a randomly chosen respondent being very confident in their detection of security breaches in the past year, we will divide the number of respondents who are very confident by the total number of respondents: \(P(\text{Very Confident}) = \frac{\text{Number of Respondents with Very Confident}}{\text{Total Number of Respondents}}\) \(P(\text{Very Confident}) = \frac{21}{106}\) So, the probabilities are: a. The probability that a respondent in the survey selected at random had little or no confidence that he or she had detected all significant security breaches in the past year is \(\frac{29}{106}\). b. The probability that a respondent in the survey selected at random was very confident that he or she had detected all significant security breaches in the past year is \(\frac{21}{106}\).

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Key Concepts

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

Probability Calculation
Understanding the probability of an event involves calculating the likelihood of that event occurring within a defined set of circumstances. It lays the foundation for making predictions based on statistical data, and is essential for a variety of fields including but not limited to, insurance, finance, security, and healthcare.

In our case, the probability calculation involves determining how likely it is for a respondent from the surveyed group to express a certain level of confidence in having detected all significant security breaches. We start by identifying the total number of outcomes, which is the total number of respondents. Then, for each scenario, we find the number of favorable outcomes (e.g., the number of respondents who are 'very confident' or have 'little or no confidence').

To calculate the specific probability, we use the formula: \(P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). When applying this formula to our survey data, for instance, the probability of selecting someone who is 'very confident' is computed as \(P(\text{Very Confident}) = \frac{21}{106}\), simplifying the concept into an accessible numerical representation of confidence levels among IT professionals regarding security breach detection.
Security Breaches Detection Confidence
In the realm of data security, the confidence with which a professional can assert that all significant security breaches have been detected is crucial. It provides an insight into the effectiveness of an organization's security measures and readiness to respond to cyber threats.

In this exercise, we analyze the confidence levels among senior IT professionals by using survey data. By categorizing confidence levels and calculating the probability of each category, we offer a numeric perspective on detection confidence. This information becomes extremely powerful when making strategic decisions about improving security protocols or investing in new security technologies.

Improving the survey's construction might involve adding granularity to understand the factors undermining confidence, or perhaps adding a timeframe to determine if confidence levels fluctuate during different periods. Such details could provide more insightful data that may lead to targeted improvements in security breach detection systems.
Survey Data Analysis
Survey data analysis plays a key role in interpreting responses from participants to draw meaningful conclusions. It encompasses organizing, summarizing, and interpreting data to uncover patterns or trends. In our exercise, the survey data uncovers varying degrees of confidence among professionals concerning security breach detection.

For effective survey data analysis, it's essential to clearly define the categories and ensure questions are unbiased to prevent skewed data. In our case, respondents were classified into four distinct confidence levels. Applying probability calculations, as demonstrated, allows for a quantification of the abstract concept of 'confidence', turning subjective opinions into measurable statistics.

By delving into the intricacies of such data, organizations can benchmark their security practices against industry standards and identify areas needing improvement.

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Most popular questions from this chapter

Data compiled by the Department of Justice on the number of people arrested in a certain year for serious crimes (murder, forcible rape, robbery, and so on) revealed that \(89 \%\) were male and \(11 \%\) were female. Of the males, \(30 \%\) were under 18 , whereas \(27 \%\) of the females arrested were under 18 . a. What is the probability that a person arrested for a serious crime in that year was under 18 ? b. If a person arrested for a serious crime in that year is known to be under 18 , what is the probability that the person is female?

The chief loan officer of La Crosse Home Mortgage Company summarized the housing loans extended by the company in 2007 according to type and term of the loan. Her list shows that \(70 \%\) of the loans were fixed-rate mortgages \((F), 25 \%\) were adjustable-rate mortgages \((A)\), and \(5 \%\) belong to some other category \((O)\) (mostly second trust-deed loans and loans extended under the graduated payment plan). Of the fixed-rate mortgages, \(80 \%\) were 30 -yr loans and \(20 \%\) were 15 -yr loans; of the adjustable-rate mortgages, \(40 \%\) were 30 -yr loans and \(60 \%\) were 15 -yr loans; finally, of the other loans extended, \(30 \%\) were 20 -yr loans, \(60 \%\) were 10 -yr loans, and \(10 \%\) were for a term of 5 yr or less. a. Draw a tree diagram representing these data. b. What is the probability that a home loan extended by La Crosse has an adjustable rate and is for a term of 15 yr? c. What is the probability that a home loan cxtended by La Crosse is for a term of 15 yr?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Let \(S=\left\\{s_{1}, s_{2}, \ldots, s_{n}\right\\}\) be a uniform sample space for an experiment. If \(n \geq 5\) and \(E=\left\\{s_{1}, s_{2}, s_{5}\right\\}\), then \(P(E)=3 / n\).

In a survey to determine the opinions of Americans on health insurers, 400 baby boomers and 600 pre-boomers were asked this question: Do you believe that insurers are very responsible for high health costs? Of the baby boomers, 212 answered in the affirmative, whereas 198 of the pre-boomers answered in the affirmative. If a respondent chosen at random from those surveyed answered the question in the affirmative, what is the probability that he or she is a baby boomer? A pre-boomer?

Propuct Reuasiumr The proprietor of Cunningham's Hardware Store has decided to install floodlights on the premises as a measure against vandalism and theft. If the probability is \(.01\) that a certain brand of floodlight will burn out within a year, find the minimum number of floodlights that must be installed to ensure that the probability that at least one of them will remain functional for the whole year is at least .99999. (Assume that the floodlights operate independently.)

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