Hello
Q - 1 -
Given that, in the normal distribution,
Mean = | 45.000 |
Standard Deviation = | 6.000 |
Q - 2 -
Given that:
Mean = | 30.000 |
Standard Deviation = | 4.000 |
95% chance means p value of 0.95 and z-score of 1.64
Hence, required X = (1.64*4)+ 30 = 36.5794 min
Hence, you should leave latest by 7:30 - 36.58 min = 06:53 (Or to be precise 06:53:25)
Thanks!
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