In a certain store, there is a 0.05 probability that the scanned price in the ba
ID: 3174312 • Letter: I
Question
In a certain store, there is a 0.05 probability that the scanned price in the bar code scanner will not match the advertised price The cashier scans 886 items. (a-1) What is the expected number of mismatches? (Round your answer to the nearest whole number.) (a-2) What is the standard deviation? (Use your rounded number for the expected number of mismatches for the calculation of standard deviation. Round your final answer to 4 decimal places.) (b) What is the probability of at least 34 mismatches? (Round the z-value to 2 decimal places. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimal places.) (c) What is the probability of more than 52 mismatches? (Round the z-value to 2 decimal places. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimal places.)Explanation / Answer
Normal Approximation to Binomial Distribution
a.
Mean ( np ) =886 * 0.05 = 44.3
b.
Standard Deviation ( npq )= 886*0.05*0.95 = 6.4873
c.
P(X > 34) = (34-44.3)/6.4873
= -10.3/6.4873 = -1.5877
= P ( Z >-1.588) From Standard Normal Table
= 0.9441
d.
P(X > 52) = (52-44.3)/6.4873
= 7.7/6.4873 = 1.1869
= P ( Z >1.187) From Standard Normal Table
= 0.1176