Arandom 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?

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

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3