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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
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