In a certain store, there is a 0.02 probability that the scanned price in the ba
ID: 3235609 • Letter: I
Question
In a certain store, there is a 0.02 probability that the scanned price in the bar code scanner will not match the advertised price. The cashier scans 751 items.
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.)
What is the probability of at least 6 mismatches? (Round the z-value to 2 decimal places. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimal places.)
What is the 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.)
In a certain store, there is a 0.02 probability that the scanned price in the bar code scanner will not match the advertised price. The cashier scans 751 items.
Explanation / Answer
n = 751 , p = 0.02 , q = 1 - p = 1 - 0.02 = 0.98
a1)
mean = n p = 751 * 0.02 = 15.02
a2)
std. deviation = sqrt(npq)
= sqrt(751 * 0.02 * 0.98)
= 3.8366
b)
P(X > 6)
z = ( x -mean) / s
= ( 6 - 15.02) / 3.8366
= -2.351
P(X > 6) = P( z > -2.351) = 0.9906
c)
P(X > 22)
z = ( x -mean) / s
= ( 22 - 15.02) / 3.8366
= -3.3936
P(X > 22) = P( z > -3.3936) = 0.0344