This is a review for my upcoming test so I need very clear i

This is a review for my upcoming test, so I need very clear instructions on how to do this.

Thank you for your assistnace.

Please do NOT, just give me the answer, explain what each variable is etc.

Thanks again,

7. At an online purchase at this store, store, 74% of customers pay with a credit card of the next 83 customers who make a a. . what is the probability that more than 70 people pay with a with a credit card? b. what is the probability that at most 40 people pay with a credit card? c. . how many of the 83 customers do you expect to pay with a credit card? what is the probability that the number of customers who pay with a credit card is within one standard deviation of the expected number? d.

Solution

Normal Distribution
Mean ( np ) =61.42
Standard Deviation ( npq )= 83*0.74*0.26 = 3.9961
Normal Distribution = Z= X- u / sd                   
a)
P(X > 70) = (70-61.42)/3.9961
= 8.58/3.9961 = 2.1471
= P ( Z >2.147) From Standard Normal Table
= 0.0159                  
b)
P(X > 40) = (40-61.42)/3.9961
= -21.42/3.9961 = -5.3602
= P ( Z >-5.36) From Standard Normal Table
= 1                  
p(X = 40) = 1 - P(X > 40) = 1 - 1 = 0

c)
Expect t o pay credit card = Mean ( np ) =61.42 ~ 62