Explanation Given: 6 % of the ball bearings shipped are rejected because they fail to meet the tolerance requirement. Approach: Write the expressions to obtaining the mean and standard deviation. μ = n p …… ( 1 ) σ = n p q …… ( 2 ) Write the equation of probability P ( X > a ) = P ( Z > a − μ σ ) …… ( 3 ) Here, n is the number of the trials, p is the probability of success and q is the probability of failure, μ is the mean, σ is the standard deviation. Calculation: Consider the equation of mean. Substitute 200 for n , 0.06 for p in the equation ( 1 ) . μ = 200 × 0.06 = 12 Substitute 200 for n , 0.06 for p and 0.94 for q in the equation ( 2 ) . σ = 200 ×

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Author: Tan

Publisher: CENGAGE L

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Chapter 8.6, Problem 20E

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The probability that a shipment of $200$ ball bearings contains more than $10$ rejects.

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Chapter 8.6, Problem 19E

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Ch. 8.1 - What is a random variable? Give an example.Ch. 8.1 - Prob. 2CQ Ch. 8.1 - Prob. 3CQ Ch. 8.1 - Three balls are selected at random without...Ch. 8.1 - 2.A coin is tossed four times. Let the random...Ch. 8.1 - A die is rolled repeatedly until a 6 falls...Ch. 8.1 - Cards are selected one at a time without...Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The probability that a shipment of 200 ball bearings contains more than 10 rejects.

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The probability that a shipment of 200<m:math><m:mrow><m:mn>200</m:mn></m:mrow></m:math> ball bearings contains more than 10<m:math><m:mrow><m:mn>10</m:mn></m:mrow></m:math> rejects.

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The probability that a shipment of $200$ ball bearings contains more than $10$ rejects.