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29 March, 05:42

A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was computed based on the sample data. The confidence interval for the mean is: (107.62, 129.75). What is the z-value that was used in the computation. Round your z-value to 2 decimal places. Select one: A. 1.45 B. 1.55 C. 1.73 D. 1.88 E. 1.96 E-mail fraud (phishing) is becoming an increasing problem for users of the internet. Suppose that 70% of all internet users experience e-mail fraud. If 50 internet users were randomly selected, what is the probability that no more than 25 were victims of e-mail fraud? Answer:

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  1. 29 March, 07:06
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    Step-by-step explanation:

    1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

    The correct option is D

    2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

    p = 70/100 = 0.7

    q = 1 - p = 1 - 0.7 = 0.3

    n = number of selected users = 50

    Mean, u = np = 50*0.7 = 35

    Standard deviation, u = √npq = √50*0.7*0.3 = 3.24

    x = number of internet users

    The formula for normal distribution is expressed as

    z = (x - u) / s

    We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

    P (x lesser than or equal to 25)

    The z value will be

    z = (25 - 35) / 3.24 = - 10/3.24 = - 3.09

    Looking at the normal distribution table, the corresponding z score is 0.001

    P (x lesser than or equal to 25) = 0.001
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