Question

Liam is a professional darts player who can throw a bullseye 70% of the time.

If he throws a dart 250 times, what is the probability he hits a bulls eye:

a.) At least 185 times?

b.) No more than 180 times?

c.) between 160 and 185 times (including 160 and 185)?

Use the Normal Approximation to the Binomial distribution to answer this question.

2. 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 51?

c.) between 35 and 49 (including 35 and 49)?

Use the Normal Approximation to the Binomial distribution to answer this question.

Answer 1

Question 1

Using Normal Approximation to Binomial
Mean = n * P = ( 250 * 0.7 ) = 175
Variance = n * P * Q = ( 250 * 0.7 * 0.3 ) = 52.5
Standard deviation = √(variance) = √(52.5) = 7.2457

Part a)
P ( X >= 185 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 185 - 0.5 ) =P ( X > 184.5 )

X ~ N ( µ = 175 , σ = 7.2457 )
P ( X > 184.5 ) = 1 - P ( X < 184.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 184.5 - 175 ) / 7.2457
Z = 1.31
P ( ( X - µ ) / σ ) > ( 184.5 - 175 ) / 7.2457 )
P ( Z > 1.31 )
P ( X > 184.5 ) = 1 - P ( Z < 1.31 )
P ( X > 184.5 ) = 1 - 0.9049
P ( X > 184.5 ) = 0.0951

Part b)
P ( X <= 180 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 180 + 0.5 ) = P ( X < 180.5 )

X ~ N ( µ = 175 , σ = 7.2457 )
P ( X < 180.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 180.5 - 175 ) / 7.2457
Z = 0.76
P ( ( X - µ ) / σ ) < ( 180.5 - 175 ) / 7.2457 )
P ( X < 180.5 ) = P ( Z < 0.76 )
P ( X < 180.5 ) = 0.7764

Part c)
P ( 160 <= X <= 185 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 160 - 0.5 < X < 185 + 0.5 ) = P ( 159.5 < X < 185.5 )

X ~ N ( µ = 175 , σ = 7.2457 )
P ( 159.5 < X < 185.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 159.5 - 175 ) / 7.2457
Z = -2.14
Z = ( 185.5 - 175 ) / 7.2457
Z = 1.45
P ( -2.14 < Z < 1.45 )
P ( 159.5 < X < 185.5 ) = P ( Z < 1.45 ) - P ( Z < -2.14 )
P ( 159.5 < X < 185.5 ) = 0.9265 - 0.0162
P ( 159.5 < X < 185.5 ) = 0.9103

Question 2

Using Normal Approximation to Binomial
Mean = n * P = ( 150 * 0.28 ) = 42
Variance = n * P * Q = ( 150 * 0.28 * 0.72 ) = 30.24
Standard deviation = √(variance) = √(30.24) = 5.4991

Part a)
P ( X >= 30 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 30 - 0.5 ) =P ( X > 29.5 )

X ~ N ( µ = 42 , σ = 5.4991 )
P ( X > 29.5 ) = 1 - P ( X < 29.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 29.5 - 42 ) / 5.4991
Z = -2.27
P ( ( X - µ ) / σ ) > ( 29.5 - 42 ) / 5.4991 )
P ( Z > -2.27 )
P ( X > 29.5 ) = 1 - P ( Z < -2.27 )
P ( X > 29.5 ) = 1 - 0.0116
P ( X > 29.5 ) = 0.9884

Part b)
P ( X <= 51 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 51 + 0.5 ) = P ( X < 51.5 )

X ~ N ( µ = 42 , σ = 5.4991 )
P ( X < 51.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 51.5 - 42 ) / 5.4991
Z = 1.73
P ( ( X - µ ) / σ ) < ( 51.5 - 42 ) / 5.4991 )
P ( X < 51.5 ) = P ( Z < 1.73 )
P ( X < 51.5 ) = 0.9582

Part c)
P ( 35 <= X <= 49 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 35 - 0.5 < X < 49 + 0.5 ) = P ( 34.5 < X < 49.5 )

X ~ N ( µ = 42 , σ = 5.4991 )
P ( 34.5 < X < 49.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 34.5 - 42 ) / 5.4991
Z = -1.36
Z = ( 49.5 - 42 ) / 5.4991
Z = 1.36
P ( -1.36 < Z < 1.36 )
P ( 34.5 < X < 49.5 ) = P ( Z < 1.36 ) - P ( Z < -1.36 )
P ( 34.5 < X < 49.5 ) = 0.9131 - 0.0869
P ( 34.5 < X < 49.5 ) = 0.8262