Question

In a study done in 2015, it was estimated that about 45% of the world’s population used some form of online social media. The average time spent per day by social media users was found to be 2.22 hours per day, with a standard deviation of 0.425 hours. Assume the distribution of this data to be normal.

a) What is the probability that a randomly chosen social media user spends more than 2.15 hours per day on social media?

b) What is the probability that the average time spent by a randomly selected group of 22 social media users is at most 2.15 hours?

c) What is the probability that the average time spent by a randomly selected group of 22 social media users is between 2.15 hours and 2.39 hours?

Answer 1

Solution :

Given that ,

mean = = 2.22

standard deviation = = 0.425

a) P(x >2.15 ) = 1 - p( x< 2.15)

=1- p [(x - ) / < (2.15-2.22) /0.425 ]

=1- P(z < -0.16 )

= 1 - 0.4364 = 0.5636

probability = 0.5636

b)

n = 22

= = 2.22  

= / n = 0.425/ 22 = 0.0906

P( 2.15) = P(( - ) / (2.15 -2.22) /0.0906 )

= P(z -0.77)

= 0.2206

probability = 0.2206

c)

P(2.15 <   < 2.39 )  

= P[(2.15-2.22) /0.0906< ( - ) / < (2.39-2.22) /0.0906 )]

= P( -0.77< Z < 1.88 )

= P(Z < 1.88) - P(Z <-0.77 )

= 0.9699 - 0.2206 = 0.7493

probability = 0.7493