Question

A recent study has shown that 28% of 18-34 year olds check their Facebook/Instagram feeds before getting out of bed in the morning,

If we sampled a group of 150 18-34 year olds, what is the probability that the number of them who checked their social media before getting out of bed is:

a.) At least 31?

b.) No more than 52?

c.) between 36 and 44 (including 36 and 44)?

Answer 1

p= 0.28 , q=1-p= 1-0.28= 0.72

n= 150

MEAN= np= 150*0.28= 42

VARIANCE= npq= 150*0.28*0.72=30.24

SD= sqrt(30.24)= 5.499

a] P(X> 31)

P ( X>31 )=P ( X−μ>31−42 )=P ((X−μ)/σ>(31−42)/5.499)

Since Z=(x−μ)/σ and (31−42)/5.499=−2 we have:

P ( X>31 )=P ( Z>−2 )

Use the standard normal table to conclude that:

P (Z>−2)=0.9772

B] P(X52)

P ( X<52 )=P ( X−μ<52−42 )=P ((X−μ)/σ<(52−42)/5.499)

Since (x−μ)/σ=Z and (52−42)/5.499=1.82 we have:

P (X<52)=P (Z<1.82)

Use the standard normal table to conclude that:

P (Z<1.82)=0.9656

C] P ( 36<X<44 )=P ( 36−42< X−μ<44−42 )=P ((36−42)/5.499<(X−μ)/σ<(44−42)/5.499)

Since Z=(x−μ)/σ , (36−42)/5.499=−1.09 and (44−42)/5.499=0.36 we have:

P ( 36<X<44 )=P ( −1.09<Z<0.36 )

Use the standard normal table to conclude that:

P ( −1.09<Z<0.36 )=0.5027