Question

The time it takes a random person to get a haircut is Normally Distributed, with an average of 23.8 minutes and a standard deviation of 5 minutes. Assume that different people have independent times of getting their hair cut. Find the probability that, if there are four customers in a row (with no gaps in between), they will all be finished getting their hair cut in 1.5 hours (altogether) or less.

Answer #1

X_i be the time taken by the i th person to get a haircut

X_i follows N( 23.8, 25 )

mean( X_i ) = 23.8

SE( X_i ) = 5

S be the sum of X_i , S = X1 + X2 + X3 + X4

S follows N( 23.8*4, 25*4*4 ) or S follows N( 95.2, 400 )

mean( S ) = 95.2 minutes

SE ( S ) = 20 minutes

probability of four customers in a row (with no gaps in between), they will all be finished getting their hair cut in 1.5 hours (altogether) or less = probability of four customers in a row (with no gaps in between), they will all be finished getting their hair cut in 90 minutes (altogether) or less = P[ S < 90 ]

P[ S < 90 ] = P[ ( S - mean(S) )/SE(S) < ( 90 - mean(S) )/SE(S) ] = P[ Z < ( 90 - 95.2 )/20 ] = P[ Z < -0.26 ] = 0.397

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