A table showing Number, Value, and Total Value. The first row shows Pennies, with the entries, p, 0.01, and 0.01 p. The second r

Answer: 942

Detailed explanation: this was quite challenging!

Answer:

The chance of completing the entire package installation in under 12 minutes is 0.1271.

Step-by-step explanation:

We define X as a normal distribution representing the time taken in seconds to install the software. According to the Central Limit Theorem, X is approximately normal, where the mean is 15 and variance is 15, giving a standard deviation of √15 = 3.873.

To find the probability of the total installation lasting less than 12 minutes, which equals 720 seconds, each installation should average under 720/68 = 10.5882 seconds. Thus, we seek the probability that X is less than 10.5882. To do this, we will apply W, the standard deviation value of X, calculated via the formula provided.

Utilizing , we reference the cumulative distribution function of the standard normal variable W, with values found in the attached file.

Given the symmetry of the standard normal distribution density function, we ascertain .

Consequently, the probability that the installation process for the entire package is completed within 12 minutes is 0.1271.

Answer:

3876

Step-by-step explanation:

We have the following data:

Members of the fraternity = 20

Their attendance at five parties is organized into groups of four.

This indicates that each group consists of 4 individuals.

At least one brother will be present at exactly one of the gatherings. (The brothers are indistinguishable).

Thus, precisely one brother at a gathering leaves (20 - 1) = 19, due to their indistinct nature.

Group sizes are in fours.

Calculating 19C4:

From: nCr = n! /(n-r)! r!

19C4 = 19! / (19 - 4)! 4!

= 19! / 15! 4!

= (19 * 18 * 17 * 16) / (4 * 3 * 2)

= 93024 / 24

= 3876