Answer to Question #121008 in Statistics and Probability for Grace

Question

After reviewing previous loan records, the credit manager of a bank determines that the data follows a normal distribution. The debts have a mean of $20 000 and the probability that the loss could be greater than $25 000 or less than $15 000 is 0.418. Determine the standard deviation of the data to the nearest hundred dollars.


A company packages rice into 10 kg bags. The machine that fills the bags can be calibrated to fill to any specified mean with a standard deviation of 0.09 kg. Any bags that weigh less than 10 kg cannot be sold and must be refilled. To what mean value, to the nearest hundredth of a kilogram, should the machine be set if the company does not want to refill more than 1.5% of the bags?

Answer 1

1) $P(X \leqslant 15000) = \Phi \left(\frac{15000 - 20000}{\sigma} \right) = \frac{0.418}{2} \Leftrightarrow \\ \frac{-5000}{\sigma} = -0.81 \\ \sigma = \frac{5000}{0.81}=6200$

2) P=0.015, so z(0.015)=-2.17

-2.17=(10-mean)/0.09

mean=10+0.09*2.17=10.20 kg