Question

The length of an instant message conversation is normally distributed with a mean of 5 minutes and a standard deviation of .7 minutes. What is the probability that a conversation lasts longer than 6 minutes?

Answer 1

The z-score for 6 minutes is:

`z=(X-\mu)/(\sigma)=(6-5)/(0.7)=1.429`
Reference to a standard normal distribution table gives the result:
P(length > 6) = 0.077

Answer 2

The correct answer is:

0.0764

Step-by-step explanation:

We will use a z-score to answer this. The formula for a z-score is

`z=(X-\mu)/(\sigma)`, where μ is the mean and σ is the standard deviation. Using our information in this problem, we have

`z=(6-5)/(0.7)=(1)/(0.7)=1.43`

Using a z-table, we look up the z-score 1.43. The area under the curve to the left of, or less than, this value is 0.9236. This means the probability that the time is greater than this is 1-0.9236 = 0.0764.