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 known that 77% of all new products introduced in grocery stores fail (are taken off the market) within 2 years. If a grocery store chain introduces 60 new products, find the following probabilities. (Round your answers to four decimal places.)

(a) within 2 years 47 or more fail


(b) within 2 years 58 or fewer fail


(c) within 2 years 15 or more succeed


(d) within 2 years fewer than 10 succeed

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 60 * 0.77 ) = 46.2
Variance = n * P * Q = ( 60 * 0.77 * 0.23 ) = 10.626
Standard deviation = √(variance) = √(10.626) = 3.2598

Part a)
P ( X <= 58 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 58 + 0.5 ) = P ( X < 58.5 )

X ~ N ( µ = 46.2 , σ = 3.2598 )
P ( X < 58.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 58.5 - 46.2 ) / 3.2598
Z = 3.77
P ( ( X - µ ) / σ ) < ( 58.5 - 46.2 ) / 3.2598 )
P ( X < 58.5 ) = P ( Z < 3.77 )
P ( X < 58.5 ) = 0.9999

Part b)
P ( X >= 47 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 47 - 0.5 ) =P ( X > 46.5 )

X ~ N ( µ = 46.2 , σ = 3.2598 )
P ( X > 46.5 ) = 1 - P ( X < 46.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 46.5 - 46.2 ) / 3.2598
Z = 0.09
P ( ( X - µ ) / σ ) > ( 46.5 - 46.2 ) / 3.2598 )
P ( Z > 0.09 )
P ( X > 46.5 ) = 1 - P ( Z < 0.09 )
P ( X > 46.5 ) = 1 - 0.5359
P ( X > 46.5 ) = 0.4641

Part c)
Using Normal Approximation to Binomial
Mean = n * P = ( 60 * 0.23 ) = 13.8
Variance = n * P * Q = ( 60 * 0.23 * 0.77 ) = 10.626
Standard deviation = √(variance) = √(10.626) = 3.2598

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

X ~ N ( µ = 13.8 , σ = 3.2598 )
P ( X > 14.5 ) = 1 - P ( X < 14.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 14.5 - 13.8 ) / 3.2598
Z = 0.21
P ( ( X - µ ) / σ ) > ( 14.5 - 13.8 ) / 3.2598 )
P ( Z > 0.21 )
P ( X > 14.5 ) = 1 - P ( Z < 0.21 )
P ( X > 14.5 ) = 1 - 0.5832
P ( X > 14.5 ) = 0.4168

Part d)
P ( X < 10 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 10 - 0.5 ) = P ( X < 9.5 )

X ~ N ( µ = 13.8 , σ = 3.2598 )
P ( X < 9.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 9.5 - 13.8 ) / 3.2598
Z = -1.32
P ( ( X - µ ) / σ ) < ( 9.5 - 13.8 ) / 3.2598 )
P ( X < 9.5 ) = P ( Z < -1.32 )
P ( X < 9.5 ) = 0.0934