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Your team has allocated 30 hours per month for dealing with customer support. The time to service a customer is a normally di

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Answer #1

The time to service a customer X_i has normal distribution. X_i\sim N(40,7^2) .

\sum_{i=1}^{35}X_i\sim N(40\times 35,35\times 7^2)=N(1400,41.4126^2)

The required probability is

P\left (\sum_{i=1}^{35}X_i>30\times 60 \right )=P\left (\sum_{i=1}^{35}X_i>1800 \right )\\ P\left (\sum_{i=1}^{35}X_i>30\times 60 \right )=P\left (Z>\frac{1800 -1400}{41.4126}\right )\\ P\left (\sum_{i=1}^{35}X_i>30\times 60 \right )=P\left (Z>9.6588961\right )\\ P\left (\sum_{i=1}^{35}X_i>30\times 60 \right )=\Phi \left (-9.6588961\right )\\ P\left (\sum_{i=1}^{35}X_i>30\times 60 \right )=0.0000

We need to find T such that

P\left (\sum_{i=1}^{35}X_i>T\right )<0.0001\\ P\left ( Z>\frac{T-1400}{41.4126} \right )<0.0001\\ \Phi \left ( -\frac{T-1400}{41.4126} \right )<0.0001\\ -\frac{T-1400}{41.4126}<\Phi ^{-1}\left (0.0001 \right )\\ -\frac{T-1400}{41.4126}<-3.719016\\ \frac{T-1400}{41.4126}>3.719016\\ T>1554.0141\\ T>1554.0141/60=25.9 \textup{ hours}

At least {\color{Blue} 25.9 \textup{ hours}} per month has to be allocated.

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