Question

PROBLEM 3 Assume that the time people spend on a particular social media website on any given day is normally distributed with a mean of 450 seconds and a standard deviation of 100 seconds. a) What percentage of people will spend between 400 and 500 seconds on the site on any given day? b) What percentage of people will spend more than 10 minutes on the site on any given day? c) The website will send targeted reminders to anyone who spends less than 5 minutes on the site on any given day. What percentage of people will get a reminder?

Please use Bayes theorem and show steps. thank you!

Answer 1

Mean, $\mu$ = 450 seconds

SD, $\sigma$ = 100 seconds

Let X denote the time spend on a particular social media website by a person

Bayes Theorem cannot be used for this question as this a direct question of normal distribution

We can solve it using Z value for Normal distribution

$Z = \frac{X - \mu}{\sigma}$

(a) The required probability = P(400 < X < 500)

= P(-0.5 < Z < 0.5)

= 0.383

Thus, 38.3% of people will spend between 400 and 500 seconds on the site on any given day

(b) The required probability = P(X > 600)

= P(Z > 1.5) = 0.0668

Thus, approximately 6.68% of people will spend more than 10 minutes on the site on any given day

(c) The required probability = P(X < 300)

= P(Z < -1.5)

= 0.0668

This, approximately 6.68% of people will get a reminder