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First Data Corp. records indicate that in 2005 43% of adult email users received a phishing email. A phishing email replicates an authentic site for the purpose of stealing personal information such as account numbers and passwords. Suppose a random sample of 800 adults will be surveyed on whether they have received phishing emails to determine recent trends. Use this information to answer questions 1 through 3.
1. Give the mean and standard deviation of the sample distribution of hatp. Give your answers in decimal form. (Example: 0.398)
Mean (Round to two decimal places) =
Standard deviation (Round to three decimal places) =
Determine the probability that more than 45% of those in a survey would have received a phishing email.
2. What is the z-score for a 45% probability? (round to three decimal places, Example: 3.456)
z =
3. What is the probability that more than 45% of those in a survey would have received a phishing email? Give your answer in decimal form, rounding to three decimal places. (Example: 0.398)

Respuesta :

Answer:

1. Mean = 0.43, standard deviation = 0.0175.

2. [tex]Z = 1.14[/tex]

3. There is a 12.71% probability that more than 45% of those in a survey would have received a phishing email.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

1. Give the mean and standard deviation of the sample distribution of hatp

First Data Corp. records indicate that in 2005 43% of adult email users received a phishing email. This means that the Mean is 0.43. So [tex]\mu = 0.43[/tex]

The standard deviation is [tex]\sqrt{\frac{0.43*0.57}{800}} = 0.0175. So [tex]\sigma = 0.0175[/tex].

2. What is the z-score for a 45% probability?

This is the value of Z when [tex]X = 0.45[/tex].

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.45 - 0.43}{0.0175}[/tex]

[tex]Z = 1.14[/tex]

3. What is the probability that more than 45% of those in a survey would have received a phishing email

This probability is 1 subtracted by the pvalue of Z when [tex]X = 0.45[/tex].

For this, we have from 2. that [tex]Z = 1.14[/tex], that has a pvalue of 0.8729.

This means that there is a 1-0.8729 = 0.1271 = 12.71% probability that more than 45% of those in a survey would have received a phishing email.

MrRoyal

Probabilities are used to determine the chances of events

  • The mean is 344
  • The standard deviation is 0.018
  • The probability that more than 45% of those in the survey would have received a phishing email is 0.134

The given parameters are:

n = 800 --- the sample size

p = 0.43 -- the proportion of adults

The mean is calculated as:

[tex]E(x) = np[/tex]

So, we have:

[tex]E(x) = 800 * 0.43[/tex]

[tex]E(x) = 344[/tex]

The standard deviation is calculated as:

[tex]\sigma= \sqrt{\frac{p * (1 - p)}{n}}[/tex]

So, we have:

[tex]\sigma= \sqrt{\frac{0.43 * (1 - 0.43)}{800}}[/tex]

[tex]\sigma= 0.018[/tex]

To calculate the probability that more than 45% of those in the survey would have received a phishing email, we start by calculating the z score

[tex]z = \frac{x - \mu}{\sigma}[/tex]

So, we have:

[tex]z = \frac{0.45 - 0.43}{0.018}[/tex]

[tex]z = 1.11[/tex]

The probability is then calculated as:

[tex]P(x > 45\%) = P(z > 1.11)[/tex]

[tex]P(x > 45\%) = 0.134[/tex]

Hence, the probability that more than 45% of those in the survey would have received a phishing email is 0.134

Read more about normal distributions at:

https://brainly.com/question/25638875

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