Question

The number of hits per day on the Web site of a tool company is normally distributed with a mean of 570 and a standard deviation of 100 . What proportion of days have more than 680 hits?

Answer 1

Approximately 13.57% of days have more than 680 hits on the website, calculated using the z-score and standard normal distribution table.

Step-by-step explanation:

To find the proportion of days with more than 680 hits on the website, we apply the concept of the normal distribution. The mean is 570 and the standard deviation is 100. To determine the number of standard deviations 680 is from the mean, we calculate the z-score.

The formula for the z-score is:

Z = (X - μ) / σ

Where X is the value of interest (680), μ is the mean (570), and σ is the standard deviation (100).

Substituting the given values:

Z = (680 - 570) / 100 = 1.1

Now, using a standard normal distribution table, we look up the z-score of 1.1 to find the proportion of values that lie below 680. The table gives us a value of approximately 0.8643, which means 86.43% of days have fewer than 680 hits.

To find the proportion of days with more than 680 hits, we subtract this value from 1:

1 - 0.8643 = 0.1357

Therefore, approximately 13.57% of days have more than 680 hits.