Answer to Question #209204 in Statistics and Probability for Novella

Question

Q.1 Global financial institution transfers a large data file every evening form offices around the world to its London headquarters. Once the file is received, it must be cleaned and partitioned before being stored in the company’s data warehouse. Each file is the same size and the time required to transfer, clean, and partition a file is normally distributed, with a mean of 1.5 hours and a standard deviation of 15 minutes.

a. if one file is selected at random, what is the probability that it will take longer than 1 hour and 55 minutes to transfer, clean, and partition the file?

b. If a manager must be present until 85% of the files are transferred, cleaned, and partitioned, how long will the manager need to be there?

c. What percentage of the data files will take between 63 minutes and 110 minutes to be transferred, cleaned, and partitioned?

Answer 1

Let $X=$ the time required to transfer: $X\sim N(\mu, \sigma^2).$

Given $\mu=90, \sigma=15.$

a.

$$P(X>115)=1-P(X\leq 115)=$$

$$=1-P(Z\leq \dfrac{115-90}{15} )\approx1-P(Z\leq 1.666667)$$

$$\approx0.0478$$

b.

$$P(X<x)=P(Z<\dfrac{x-90}{15})=0.85$$

$$\dfrac{x-90}{15}\approx1.036433$$

$$x=105.5\ min$$

c.

$$P(63<X<110)=P(X<110)-P(X\leq 63)=$$

$$=P(Z< \dfrac{110-90}{15} )-P(Z\leq \dfrac{63-90}{15})$$

$$\approx P(Z< 1.333333)-P(Z\leq -1.8)$$

$$\approx 0.90879-0.03593=0.8729$$

$87.29\%$