Question

According to a social media​ blog, time spent on a certain social networking website has a mean of 22 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes. Complete parts​ (a) through​ (d) below.

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 21.5 and 22.5 minutes? (Round to three decimal places)

b.If you select a random sample of 25 sessions, what is the probability that the sample mean is between 21 and 22 minutes? (Round to three decimal places)

c. If you select a random sample of 144 sessions, what is the probability that the sample mean is between 21.5 and 22.5 minutes? (Round to three decimal places)

Answer 1

Solution :

Given that,

mean = = 22

standard deviation = = 4

n = 25

= 22

  ₌   / n = 4 25 = 0.8

a ) P ( 21.5 <   <  22.5 )

P (21.5 - 22 /0.8 ) < ( - /) < (22.5 - 22 /0.8 )

P( - 0.5 / 0.8 < z  < 0.5 / 0.8)

P ( - 0.62 < z < 0.62 )   

P ( Z < 0.62 ) - P ( Z < - 0.62 )

Using z table

= 0.7324 - 0.2676

= 0.4648

Probability = 0.465

b ) P ( 21 <   < 22 )

P (21 - 22 /0.8 ) < ( - /) < (22 - 22 /0.8 )

P( - 1 / 0.8 < z  < 1 / 0.8)

P ( - 1 < z < 1 )   

P ( Z < 1 ) - P ( Z < - 1 )

Using z table

= 0.8413 - 0.1587

= 0.6826

Probability = 0.683

c ) n = 144

= 22

  ₌   / n = 4 144 = 0.3333

P ( 21.5 <   <  22.5 )

P (21.5 - 22 /0.3333 ) < ( - /) < (22.5 - 22 /0.3333 )

P( - 0.5 / 0.3333 < z  < 0.5 / 0.3333)

P ( - 1.5 < z < 1.5 )   

P ( Z < 1.5 ) - P ( Z < - 1.5 )

Using z table

= 0.9332 - 0.0668

= 0.8664

Probability = 0.866