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A workshop makes tables. The cutting and sanding operations are independent. Based on historical data, under...

A workshop makes tables. The cutting and sanding operations are independent. Based on historical data, under normal conditions, 1% of cut wood for the tables, and 2% of sanded wood for the tables are defective. Assume that one table is randomly selected from a lot of cut and sanded tables. a. What is the probability that at least one of the operations (cutting and sanding) will be defective? b. What is the probability that in a production lot of 10 tables, none of the tables is defective? c. What is the expected number of defective tables in a production lot of 15 tables? What is the standard deviation?

Solutions

Expert Solution

a. Let A be the event that cut wood for the tables is

defective and B be the event that sanded wood for the table

is defective.

Thus, P(A) = 0.01 and P(B) = 0.02, P(A B) = 0.

Hence, the probability that at least one of the operations

will be defective = P(A B) = P(A) + P(B) - P(A B)

= 0.01 + 0.02 = 0.03. (Ans).

b. Let X be the random variable denoting the number of

tables which are defective in a production lot of 10 tables.

Thus, X ~ Bin(10, 0.03)

Hence, the probability that none of the tables are defective

= P(X = 0) = = 0.7374. (Ans).

c. The expected number of defective tables in a production

lot of 15 tables = 15 * 0.03 = 0.45. (Ans).

The standard deviation of the number of defective tables in

a production lot of 15 tables = = 0.6607.

(Ans).

milcah answered 1 year ago
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