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in a certain store, there is a .03 probability that the scanned price in the bar code scanner will not match the advertised price. the cashier scans 800 items. (a-1) what is the expected number of mismatches? (a-2) what is the standard deviation? (round your final answer to 4 decimal places.) (b) what is the approximate normal probability of at least 20 mismatches? (round the z-value to 2 decimals. use appendix c-2 to find probabilities. round your final answer to 4 decimal places.) (c) what is the approximate normal probability of more than 30 mismatches? (round the z-value to 2 decimals. use appendix c-2 to find probabilities. round your final answer to 4 decimal places.)

Respuesta :

(a) The required standard deviation is 4.7886

(b) The required probability is 0.8944

(c) The required probability is 0.0721

Moreover, we have probability (p) = 0.03

sample size (n) = 788

putting the value n = 788 and p = 0.03, we get

Standard deviation = \sigma = \sqrt(n*p*q)

where n=788, p = 0.03 and q = 1- p= 1-0.03 = 0.97

so, required standard deviation is 4.7886

(b) Probability of at lest 18 mismatches

putting the value of \bar{x} = 18, \mu = 24, \sigma = 4.7886

P(Z\ge18) = (18-24)/(4.7886) = -6/4.7886 = -1.25 so, P(Z\ge-1.25) = P(Z<1.25) = 0.8944. Using normal standard deviation table So, the required probability is 0.8944

(c) Probability of more than 31 mismatches

putting the value of \bar{x} = 31, \mu = 24, \sigma = 4.7886

P(Z>31) = (31-24)/(4.7886) = 7/4.7886 = 1.46

so, P(Z>1.46) = 1-P(Z<1.46) = 1- 0.9289 = 0.0721(Using normal standard deviation table) So, the required probability is 0.0721.

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