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Assume the length of waiting at supermarket is approximately normally distributed with mean 6 minutes and standard deviation 1.5 minutes.

(1) Fund the probability that waiting time is longer than 8 minutes

My way is P(z > (8-6)/1.5) = P(z > 4/3) = .0918

(2) What length of the waiting time constitutes the 99th percentile ($Φ_.99$)?

My way is

For 99th percentile, find p(z > 2.327) = .01

find (y-6)/1.5 = 2.237 <=> y = 9.4905. For 0th percentile, y should be negative infinity. However, it does not make sense since we talk about time here. We should take 0 instead.

The waiting time length is between 0 and 9.4905

(3) Find probability that waiting time lasted longer than 6 minutes given that it lasted longer than 5 minutes

My way is P(Y>6|Y>5) = P(Y>6,Y>5)/P(Y>5) = P(Y>6)/P(Y>5) = P(z > (6-6)/1.5)/P(z > (5-6)/1.5) = P(z > 0) = P(z > -2/3) = .5/(1-.2514)=.67.

Could anyone please check (1), (2), (3) for me?

(1) Fund the probability that waiting time is longer than 8 minutes

My way is P(z > (8-6)/1.5) = P(z > 4/3) = .0918

(2) What length of the waiting time constitutes the 99th percentile ($Φ_.99$)?

My way is

For 99th percentile, find p(z > 2.327) = .01

find (y-6)/1.5 = 2.237 <=> y = 9.4905. For 0th percentile, y should be negative infinity. However, it does not make sense since we talk about time here. We should take 0 instead.

The waiting time length is between 0 and 9.4905

(3) Find probability that waiting time lasted longer than 6 minutes given that it lasted longer than 5 minutes

My way is P(Y>6|Y>5) = P(Y>6,Y>5)/P(Y>5) = P(Y>6)/P(Y>5) = P(z > (6-6)/1.5)/P(z > (5-6)/1.5) = P(z > 0) = P(z > -2/3) = .5/(1-.2514)=.67.

Could anyone please check (1), (2), (3) for me?

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