Suppose Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. Assume the travel times are normally distributed with a standard deviation of 10.3 min. Determine the travel time ? such that 22.96% of the 60 days have a travel time that is at least x.
x=_____min
Solution :
Given that,
mean = = 35.6
standard deviation = = 10.3
Using standard normal table ,
P(Z > z) = 22.96 %
1 - P(Z < z) = 0.2296
P(Z < z) = 1 - 0.2296
P(Z < 0.74) = 0.7704
z = 0.74
Using z-score formula,
x = z * +
x = 0.74 * 10.3 + 35.6 = 43.22
x = 43.22 mim
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