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Determine the probability that the demand.

Demands for bluetooth headsets at a wireless phone center occur at an average of 11 per
day and a standard deviation of 5. The center has an inventory of 100 and does not
expect replenishment in the next 10 business days. Determine the probability that the
demand in the next 10 business days of phone center operation is met.

The expected demand over 10 days is 110 units. This is 10 more than inventory - an expected  shortfall of 10.   The expected demand is a sum of 10 normally distributed random variables each with σ =5.  This sum is also a normally distributed random variable with mean of 110 and  standard deviation of              5 sqrt(10) ~ 15.8.    Thus the shortfall is 10/ 15.8 ~ .6325 standard deviations.

The probability of a random variable (in this case the sum) falling .6325 or more standard deviations below the expected value (110) is given by normalcdf function. By TI-84 this is ~26.4%
You can use the Central Limit Theorem (CLT) "in reverse" here. If each business day (sample) has a normal distribution of mean 10 and standard deviation 5, then for 10 business days, we expect a normal distribution with mean 11*10=110 and standard deviation 5√(10), N(110, 5√(10)). The demand is met when X≤100, so we need the maximum allowed z-score, z=(100-110)/5√(10)≈-.6324. Now use normalcdf to find

P(Z≤-.6324)=.2635.