Answer:
Answer would be 0.00621
Step-by-step explanation:
It given that the length of a social media interaction is normally distributed with a mean of 3 minutes and standard deviation of 0.4 minutes.
The probability that a randomly selected data, X, is greater than a given value, x, is given by
[tex]P(X \ \textgreater \ x)=P \left(z\ \textgreater \ \frac{x-\mu}{\sigma} \right)=1-P \left(z\ \textless \ \frac{x-\mu}{\sigma} \right)[/tex]
The probability that an interaction lasts longer than 4 minutes is given by
[tex]P(X\ \textgreater \ 4)=P\left(X\ \textgreater \ \frac{4-3}{0.4} \right)
=P(X\ \textgreater \ 2.5)=1-P(X\ \textless \ 2.5)[/tex]
Now using normal distribution table (z table) or calculator to evaluate that
P(X< 2.5) = 0.99379
Therefore the probability that an interaction lasts longer than 4 minutes
= 1 - 0.99379 = 0.00621
