Answer:
Q1 a) 0.9545
b) 0.02275
c) 0.97725
Q2 z is approximately equal to -1.466
Step-by-step explanation:
Q1 The given information are;
The mean time to commute to work = 24.4 minutes
The standard deviation = 6.5 minutes
a) The z-score for 11.4 is given as follows;
[tex]Z=\dfrac{x-\mu }{\sigma }[/tex]
Where;
x = Observed value 11.4
μ = The mean = 24.4 minutes
σ = The standard deviation = 6.5 minutes
[tex]Z=\dfrac{11.4-24.4 }{6.5 } = -2[/tex]
The z-score for 37.4 is given as follows;
[tex]Z=\dfrac{37.4-24.4 }{6.5 } = 2[/tex]
-2 < z < 2, which gives, from the z-score table;
The probability of commute time to be between 11.4 minutes and 37.4 minutes = 0.97725 - 0.02275 = 0.9545
b) From the z-score table, the probability that the commute time to be less than 11.4 minutes = The probability at z = -2 = 0.02275
c) From the z-score table, the probability that the commute time to be greater than 37.4 minutes = The probability at z = 2 = 0.97725
Q2 The the z-score that corresponds to a commute time of 15 minutes is given as follows;
[tex]Z=\dfrac{x-\mu }{\sigma }[/tex]
[tex]Z=\dfrac{15-24.4 }{6.5 } = -\dfrac{94}{65} \approx -1.466[/tex]
