The probability for part a is 14.56%, for part b is 0, and for a part, c is 1.35%.
It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.
From the Normal Approximation to Binomial
Mean = n×P = (1100× 0.1) = 110
Variance = n×P×Q = 1100×0.1×0.9 ) = 99
Standard deviation = √(variance) = √(99) = 9.94
Condition for Normal Approximation to Binomial
n×P ≥ 10
n×(1 - P ) ≥ 10
Part a)
P ( X ≥ 121 )
P(X > n - 0.5) = P(X > 121 - 0.5) = P( X > 120.5 )
X ~ N (µ = 110 , σ = 9.94)
P (X > 120.5) = 1 - P (X < 120.5)
Z = ( X - µ ) / σ
Z = ( 120.5 - 110 ) / 9.9499
Z = 1.0553
P ( ( X - µ ) / σ ) > ( 120.5 - 110 ) / 9.9499 )
P ( Z > 1.0553 )
P ( X > 120.5 ) = 1 - P ( Z < 1.0553 )
P ( X > 120.5 ) = 1 - 0.8544
P ( X > 120.5 ) = 0.1456 ≈ 14.56%
Part b)
P ( X ≥ 165 )
P ( X > n - 0.5 ) = P ( X > 165 - 0.5 ) = P ( X > 164.5 )
X ~ N ( µ = 110 , σ = 9.9499 )
P ( X > 164.5 ) = 1 - P ( X < 164.5 )
Standardizing the value
Z = ( 164.5 - 110 ) / 9.9499
Z = 5.4774
P ( ( X - µ ) / σ ) > ( 164.5 - 110 ) / 9.9499 )
P ( Z > 5.4774 )
P ( X > 164.5 ) = 1 - P ( Z < 5.4774 )
P ( X > 164.5 ) = 0 ≈ 0%
Part c)
Sampling distribution P is approximately normal if np ≥ 10 and n (1-p) ≥ 10
n×p = 1100×0.1 = 110
n×(1 - p ) = 1100×(1 - 0.1) = 990
Mean ⇒ p = 0.1
Standard deviation = p(1 - p)/n = 0.009045
X ~ N ( µ = 0.1 , σ = 0.009045 )
P ( X < 0.08 )
Z = (0.08 - 0.1)/0.009045
Z = -2.2112
P ( ( X - µ ) / σ ) < ( 0.08 - 0.1 ) / 0.009045 )
P ( P < 0.08 ) = 0.0135 ≈ 1.35%
Thus, the probability for part a is 14.56%, for part b is 0, and for part, c is 1.35%.
Learn more about the probability here:
brainly.com/question/11234923
#SPJ1
