
Robert Z. answered 08/25/20
3100+ hours (& counting!) tutoring math -- Prealgebra to Calculus 2
P(defective) = .12. We will just call this p.
P(not defective) = .88
X = number defective in a batch of 50
a. Avery
The probability is most easily found with the appropriate distribution function on a graphing calculator.
In a batch of 50, P(X ≥ 3) = 1 - P(X ≤ 2) = 1 - binomcdf(50,.12,2) = .949
b. Bradley
To use the normal approximation, we would first find the mean and the standard deviation for the number of defective units in a batch of 50. The mean is μ = np = 50(.12) = 6.
The standard deviation is σ = √np(1-p) = √50(.12)(.88) = 2.30.
In the normal approximation, the discrete 2's, are represented by the continuous range from 1.5 to 2.5
Putting that together, we want the area under the normal distribution curve from 0 to 2.5.
The z-score for 2.5 is (2.5 - 6)/ 2.3 = -1.52
Finding this z-score in a table of the normal distribution, or using the normdcf function on the graphing calculator, we find the P(X ≤ 2) = normcdf (-999,-1.52,0,1) = .064
Finally, P(X ≥ 3) = 1 - P(X ≤ 2) = 1 - .064 = .936