In a certain store, there is a 0.02 probability that the scanned price in the ba

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 796 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 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 probability of more than 26 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

Explanation / Answer

n= 796
p = .02

So, params of binomial dist are given above

We will use them to approximate a normal distribution and solve the parts using normal distribution Z tables as below:

a-1. E(X) = np = 796*.02 = 15.92 or 16

a-2. Stdev = sqrt(npq) = sqrt(796*.02*.98) = 3.9499

b. P(X>=10) = 1- P(X<10) = 1- P(Z< 10-16 / 3.9499) = 1 - 0.0643 = .9357

c. P(X>26) = 1-P(X<=25) = 1- P(Z< 25-16 / 3.9499) = 1-P(Z<2.28) = 1-0.98865 = 0.0113

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