Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. It is estimated that 3.5% of the general population will live past their 90th birthday. In a graduating class of 765 high school seniors, find the following probabilities. (Round your answers to four decimal places.)

(a) 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

(d) more than 40 will live beyond their 90th birthday

Answer 1

Mean = n * P = ( 765 * 0.035 ) = 26.775
Variance = n * P * Q = ( 765 * 0.035 * 0.965 ) = 25.8379
Standard deviation = √(variance) = √(25.8379) = 5.0831

Part a)

P ( X >= 15 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 15 - 0.5 ) =P ( X > 14.5 )

X ~ N ( µ = 26.775 , σ = 5.0831 )
P ( X > 14.5 ) = 1 - P ( X < 14.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 14.5 - 26.775 ) / 5.0831
Z = -2.41
P ( ( X - µ ) / σ ) > ( 14.5 - 26.775 ) / 5.0831 )
P ( Z > -2.41 )
P ( X > 14.5 ) = 1 - P ( Z < -2.41 )
P ( X > 14.5 ) = 1 - 0.008
P ( X > 14.5 ) = 0.9920

Part b)

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

X ~ N ( µ = 26.775 , σ = 5.0831 )
P ( X > 29.5 ) = 1 - P ( X < 29.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 29.5 - 26.775 ) / 5.0831
Z = 0.54
P ( ( X - µ ) / σ ) > ( 29.5 - 26.775 ) / 5.0831 )
P ( Z > 0.54 )
P ( X > 29.5 ) = 1 - P ( Z < 0.54 )
P ( X > 29.5 ) = 1 - 0.7054
P ( X > 29.5 ) = 0.2946

Part c)

P ( 25 < X < 35 )
Using continuity correction
P ( n + 0.5 < X < n - 0.5 ) = P ( 25 + 0.5 < X < 35 - 0.5 ) = P ( 25.5 < X < 34.5 )

X ~ N ( µ = 26.775 , σ = 5.0831 )
P ( 25.5 < X < 34.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 25.5 - 26.775 ) / 5.0831
Z = -0.25
Z = ( 34.5 - 26.775 ) / 5.0831
Z = 1.52
P ( -0.25 < Z < 1.52 )
P ( 25.5 < X < 34.5 ) = P ( Z < 1.52 ) - P ( Z < -0.25 )
P ( 25.5 < X < 34.5 ) = 0.9357 - 0.401
P ( 25.5 < X < 34.5 ) = 0.5347

Part d)

P ( X > 40 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 40 + 0.5 ) = P ( X > 40.5 )

X ~ N ( µ = 26.775 , σ = 5.0831 )
P ( X > 40.5 ) = 1 - P ( X < 40.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 40.5 - 26.775 ) / 5.0831
Z = 2.7
P ( ( X - µ ) / σ ) > ( 40.5 - 26.775 ) / 5.0831 )
P ( Z > 2.7 )
P ( X > 40.5 ) = 1 - P ( Z < 2.7 )
P ( X > 40.5 ) = 1 - 0.9965
P ( X > 40.5 ) = 0.0035