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 30?  

b.) No more than 55?  

c.) between 40 and 49 (including 40 and 49)?

Answer 1

X ~ Bin ( n,p)

Where n = 150 , p = 0.28

Mean = n p = 150 * 0.28 = 42

Standard deviation = sqrt [ np ( 1 - p) ] = sqrt [ 150 * 0.28 ( 1 - 0.28) ] = 5.4991

Using normal approximation,

P(X < x) = P( Z < ( x - mean) / SD )

a)

With continuity correction

P(X >= 30 ) = P(X > 29.5)

P ( X > 29.5 ) = P(Z > (29.5 - 42 ) / 5.4991 )
= P ( Z > -2.27 )
= 1 - P ( Z < -2.27 )
= 1 - 0.0116
= 0.9884

b)

With continuity correction

P(X <= 55) = P(X < 55.5)

P ( ( X < 55.5 ) = P ( Z < 55.5 - 42 ) / 5.4991 )
= P ( Z < 2.45 )
P ( X < 55.5 ) = 0.9929

c)

With continuity correction

P ( 40 <= X <= 49) = P ( 39.5 < X < 49.5 )

P ( 39.5 < X < 49.5 ) = P ( Z < ( 49.5 - 42 ) / 5.4991 ) - P ( Z < ( 39.5 - 42 ) / 5.4991 )
= P ( Z < 1.36) - P ( Z < -0.45 )
= 0.9131 - 0.3264
= 0.5867