The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the variance of the waiting time is 4. Find the probability that a person will wait for between 6 and 9 minutes. Round your answer to four decimal places
Solution :
Given that ,
mean = = 5
2 =4
standard deviation = = 2
P(6< x < 9) = P[(6-5) /2 < (x - ) / < (9-5) / 2)]
= P( 0.5< Z <2 )
= P(Z < 2) - P(Z < 0.5)
Using z table
= 0.9772-0.6915
probability= 0.2857
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