1. Ms. Busybee has 10 tasks to be completed at the beginning of her day (includi

Question

1. Ms. Busybee has 10 tasks to be completed at the beginning of her day (including lunch). Each task takes a Normally distributed time with mean and standard deviation given in the table below (in minutes Task1 Mean 30 Stdev|9 4 50 10 40 20 10 60 12 40 40 50 4 60 Since each task is normally distributed, we do not have to assume that 10 is a large number of tasks. What is the probability of completing the tasks within 7 hours? If she starts her day at 8am and does nothing but the above tasks, what time should she schedule a meeting with someone if she wants to be 95% sure of completing her 10 tasks before the meeting?

Explanation / Answer

It is not given that the tasks are dependent among themselves. So, assuming independence among the tasks let us define a random variable Y which is equal to the sum of the above 10 variables mentioned in the table.

Then, Y = X1 + X2 + ... + X10.

Mean of Y, E(Y) = E(X1 ) + E(X2 ) + ... + E(X10 ) = 400.

Variance of Y, Var(Y) = Var(X1 ) + Var(X2 ) + ... + Var(X10 ) [ Due to independence ]

= 81 + 64 + 49 + 100 + 196 + 121 + 9 + 144 + 16 + 4 = 784.

Standard deviation of Y, sd(Y) = sqrt(784) = 28.

Now, the probability of completing the tasks within 7 hours means Y is taking the value 7*60=420 minutes or less,

P(Y < 420) = P((Y - 400)/28 < (420-400)/28) = P (Z < 0.7143) = 0.7625.

To find the time to be 95% sure to complete all 10 jobs, let that time be 't' minutes. Then,

P((Y - 400)/28 < (t - 400)/28) = 0.95 i.e. (t - 400)/28 = 1.645 i.e. t = 400 + 28*1.645 = 446.06 mins = 7 hrs 26 mins.

Therefore, she should schedule the meeting after (8 am + 7 hrs 26 mins) = 3:26 pm.

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