In a certain store, there is a 0.04 probability that the scanned price in the ba
ID: 3250316 • Letter: I
Question
In a certain store, there is a 0.04 probability that the scanned price in the bar code scanner will not match the advertised price. The cashier scans 900 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 28 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 45 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
There are two outcomes, success-bar code scanner not matching advertised price and failure-bar code scanner matching advertised price. There are n=900 random, independent trials, and probability of success, p is 0.04. The probability of success is constant throughout the trials. This accounts for binomial distribution. But the number of trials is large and p is small compared to it. Therefore, apply normal approximation to binomial distribution.
a-1. One can expect, mu=np=900*0.04=36 mismatches.
a-2 The standard deviation of mismatch is sqrt npq=sqrt np(1-p)=sqrt 900*0.04*(1-0.04)=5.88(ans)
b. Assume X be the r.v denoting the number of mismatches. To find required probability use the expected mismatches and standard deviation of mismatches to compute the z score using z=(x-mu)/sigma, where, X denote the raw score.
P(X>=28)=1-P(X<28)=1-P(X<=27)=1-P[Z<=(27-36)/5.88]=1-P(Z<=-1.53)=1-0.0630=0.9370 (ans)
c. P(X>45)=1-P(X<=45)=1-P[Z<=(45-36)/5.88]=1-P(Z<=1.53)=1-0.9370=0.0630 (ans)