Question

In a certain store, there is a .02 probability that the scanned price in the bar code scanner will not match the advertised price. The cashier scans 844 items.

(a-1) What is the expected number of mismatches? (Round your answer to the nearest whole number.)
  
Expected 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.)
  
Standard deviation            

(b) What is the approximate normal probability of at least 10 mismatches? (Round the z-value to 2 decimal places. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimal places.)
  
Probability            

(c) What is the approximate normal probability of more than 22 mismatches? (Round the z-value to 2 decimal places. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimal places.)
  
Probability            

Answer 1

Given that, n=844, p=0.02 a-1) The expected number of mismatches is, A =np =844*0.02 =16.88 =17 a-2) The standard deviation is, o= Vnp(1-p) = 844*0.02*(1-0.02) = 4.0672 The approximate normal probability of at least 10 mismatches is, P(XBinomias 210) = P(X.xomas > 10-0.5) = P xormai , 9.5-17 0 4 .0672) =1-PC25-1.84) = 1-0.0329 (Using the z table

= 0.9671 The approximate normal probability of more than 22 mismatches is, P(XBinomial > 22) = P(X.Normal > 22+0.5) = P( Veran 4 - 22:56-577) =1-P(251.35) (Using the z table) = 1-0.9115 = 0.0885