What is the probability that at least 18 of the 50 subscribers in your sample REOFFEND? Find the mean of the sample proportion of subscribers that REOFFENDED. Report to two decimal places

The question has some details missing and incomplete. here is the complete question ; New regulations in Canada require all Internet service providers (ISPs) to send a notice to subscribers who are downloading files illegally asking them to stop. This "notice and notice" system was already in place with Rogers Cable. That company says that prior to these regulations, 67% of its subscribers who received the notice did NOT reoffend. Consider a random sample of 50 of these Rogers subscribers who received a first notice.

a) What is the distribution (or approximate distribution) of the number X of subscribers who reoffend?

b)Find the mean of the sample proportion of subscribers that REOFFENDED. Report to two decimal places:

c)Find the standard deviation of the sample proportion of subscribers that REOFFENDED. Report to three decimal places:

d)Find the z-score for the 18 of the 50 subscribers who reoffended (report to two decimal places)

e)Find the probability that at least 18 of the 50 subscribers reoffended (report to two decimal places)

Step-by-step explanation:

a) Approximate distribution of the number X of subscribers who reoffend is NORMAL, as both np and n(1-p) are greater than 5.

b) n = 50,  67% of its subscribers who received the notice did NOT reoffend = q, therefore p = 1 - q = 1 - 0.67 = 0.3300, p = 0.3300,

here mean of distribution ; μ = np = 16.5

c) standard deviation σ =√(np(1-p)) = 3.325

d) z score =  (X-mean) / SD = (18-16.5) / 3.325=0.45

e) Probability that at least 18 of the 50 subscribers reoffended; P(X>18)  = P(Z>0.45) = 0.33

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

why can't you?

step-by-step explanation: