Asked by Jim
A popular online retailer sells a wide variety of products including books. The proportion of customers that order only books is 21%. If a random sample of 600 customers is taken, what is the probability that more than 24% order only books?
Answers
Answered by
MathGuru
Use the normal approximation to the binomial distribution.
First find mean and standard deviation.
mean = np = (600)(.21) = ?
standard deviation = √npq = √(600)(.21)(.79) = ?
Note: q = 1 - p
I'll let you finish the calculations.
Next take .24 times 600. Use that value for x in the z-score equation:
z = (x - mean)/sd
Once you calculate the z-score, look at a z-table to determine the probability. Remember that the problem is asking "more than" when you check the table.
I hope this will help get you started.
First find mean and standard deviation.
mean = np = (600)(.21) = ?
standard deviation = √npq = √(600)(.21)(.79) = ?
Note: q = 1 - p
I'll let you finish the calculations.
Next take .24 times 600. Use that value for x in the z-score equation:
z = (x - mean)/sd
Once you calculate the z-score, look at a z-table to determine the probability. Remember that the problem is asking "more than" when you check the table.
I hope this will help get you started.
Answered by
sean
0.983
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