The distribution of cash withdrawals from the automatic teller machine at a
certain bank has a mean of $500 with a standard deviation of $70. To reduce the
incentives for robbery, the bank puts money into the machine every 12 hours and
it keeps the amount deposited fairly close to the expected total withdrawals for a
12-hour period. If 100 withdrawals were expected in each 12-hour period and
each withdrawal was independent, how much should the bank put into the
machine so that the probability of running out of money was 0.05?
To find out how much money the bank should put into the machine, we need to calculate the maximum amount that is expected to be withdrawn in a 12-hour period.
Given that the mean withdrawal is $500 and the standard deviation is $70, we can use the concept of the Central Limit Theorem to approximate the maximum amount.
The Central Limit Theorem states that the sum of a large number of independent and identically distributed random variables tends towards a normal distribution, regardless of the distribution of the original variables, as long as certain conditions are met. In this case, since we have 100 independent withdrawals, we can assume that the sum of withdrawal amounts in a 12-hour period will approximately follow a normal distribution.
The mean of the sum of withdrawals in a 12-hour period can be calculated by multiplying the mean withdrawal amount by the number of withdrawals:
mean = 500 * 100 = $50,000
The standard deviation of the sum of withdrawals in a 12-hour period can be calculated by multiplying the standard deviation of a single withdrawal by the square root of the number of withdrawals:
standard deviation = 70 * sqrt(100) = $700
Now, we want to find the amount of money the bank should put into the machine such that the probability of running out of money is 0.05, or 5%. This corresponds to the lower 5th percentile of the distribution.
To calculate this, we can use a table of the standard normal distribution (z-table) or a statistical calculator. The z-score corresponding to the 5th percentile is approximately -1.645.
Now, we can use the formula for a z-score to find the maximum withdrawal amount represented by the 5th percentile:
z = (x - mean) / standard deviation
Rearranging the formula to solve for x, the maximum withdrawal amount, we get:
x = z * standard deviation + mean
x = -1.645 * $700 + $50,000
x ≈ $49,150.50
Therefore, the bank should put approximately $49,150.50 into the machine so that the probability of running out of money is 0.05.