Question

1.Tom and Joe have been told that the time it takes to prepare an order, X, is normally distribution with a mean of 90 seconds and a standard deviation of 40 second. X~N(90, 40). What is the probability that an order takes a minute or less to prepare? P(X ≤ 60)? Round to three decimal places. What is the probability that an order takes longer than three minutes to prepare, P(X ≥ 180)? Round to three decimal places.

Answer 1

Solution :

Given that ,

P(x 60 )

= P[(x - ) / (60 - 90) / 40]

= P(z - 0.75 )

Using z table,

= 0.227

P(x 180 ) = 1 - P(x   180)

= 1 - P[(x - ) / (180 - 90) / 40 ]

= 1 -  P(z 2.25 )

= 1 - 0.988

= 0.012