The amount of time Facebook users spend on the website each month is normally di

Question

The amount of time Facebook users spend on the website each month is normally distributed with the mean of 6.7 hrs, and the standard deviation of 1.8 hrs.

a) Find the probability that a facebook user spend less than four hours on a website in a month?

b)Find the probability that a facebook user spend between four hours and ten hours on a website in a month.

c) Out of 1000, facebook users, how many do you expect to spend hours between 4 to 10 on the website in a given month?

d) What is a minimum amount of time spent on Facebook in given month that would place the users in the top 10% of times?

Explanation / Answer

Population mean = 6.7, Population standard deviation = 1.8

a) We can calculate the z score corresponding to four hours as below:

z score = (observed - expected)/sd

= (4 - 6.7)/1.8

= -1.5

From the z table, we can calulate the p value for z score less than -1.5, which is 0.06681

So probability that a facebook user spends less than 4 hours on a webiste in a month is 0.06681.

b) P(4 < x < 10) = P(x < 10) - P (x < 4)

Z score corresponding to 10 hours = (10 - 6.7)/1.8 = 1.833333, corresponding p value = 0.96662

Z score corresponding to 4 hours = (4 - 6.7)/1.8 = -1.5, corresponding p value = 0.06681

So probability that a facebook user spend between 4 to 10 hours on a webiste in a month = 0.96662 - 0.06681

=0.89981

c) So if total users are 1000, we can expect 0.89981 * 1000 = 900 users approx to spend hours between 4 to 10 on the website in a given month.

d) Here we need to calculate the z score corresponding to p value of 0.90, which is equal to 1.282

1.282 = (observed - 6.7)/1.8

1.282 * 1.8 = observed - 6.7

2.3076 = observed - 6.7

observed = 9.0076

So minimum amount of time spent on Facebook in given month that would place the users in the top 10% of times is 9.0076 hours.

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