Question

1. Facebook provides a variety of statistics on this website that detailed growth in popularity of the site. On average, 28% of 18 to 35 year olds check their Facebook profile before getting out of bed in the morning. Suppose this percentage follows a normal distribution with standard deviation of 5%. A. find the probability that the percent of 18 to 35 year olds to check Facebook before getting out of bed in the morning is at least 30. B. Find the 95th percentile and express it in a sentence.

2. X-N(54,8) find the probability that x>56, EXPLAIN WHY ANSWER IS .4013 PLEASE

Answer 1

1)

A)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	28
std deviation   =σ=	5.000

probability that the percent of 18 to 35 year olds to check Facebook before getting out of bed in the morning is at least 30:

probability =

P(X>30)

=

P(Z>0.4)=

1-P(Z<0.4)=

1-0.6554=

0.3446

B)

for 95th percentile critical value of z =1.645

hence correspponding value =mean+z*std deviation =28+1.645*5= 36.23

from above one can say that there is 95% probability of having at most 36.22 % of  18 to 35 year olds that check their Facebook profile before getting out of bed in the morning.

2)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	54
std deviation   =σ=	8.000

probability =

P(X>56)

=

P(Z>0.25)=

1-P(Z<0.25)=

1-0.5987=

0.4013

( please revert for any clarification required)