A firm is manufacturing 45,000 units of nuts. The probability of having a defective nut is 0.15

Compute the given

i. The expected no. of defective nuts

ii. The standard and variance deviation of the defective nuts in a daily consignment of 45,000

Solution

Sample size n = 45,000

P(defective) = 0.15 = p

P(non defective) = 0.85 = q

i. ∴  the expected no of defective nuts

= 45,000 × 0.15 = 6,750

ii. The variance = npq

= 45000 × 0.85 × 0.15

= 5737.50

The standard deviation =  √(npq)

= √(5737.50)

= 75.74