To solve this problem we can use a z-table. First, we convert our score to a z-score using the following formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
where mu represents the mean and sigma represents the standard deviation.
Using this formula in our problem, we have:
[tex]z=\frac{6-5}{0.7}=\frac{10}{7}\approx1.429[/tex]
This z-score represents the position where the phone call is equal to 6 minutes. On a z-table, we're going to find the area between the mean and this z-score, since we want the probability that a conversation lasts longer than 6 minutes, we want the area above it. To calculate this area, we're going to subtract the value given on the z-table from 0.5.
The value on the z-table is:
Then, our probability is:
[tex]P(x>6)=0.5-0.4236\approx0.077[/tex]
The answer is 0.077.
