Maharashtra State BoardHSC Arts 12th Board Exam
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The probability that a machine will produce all bolts in a production run within specification is 0.998 - Mathematics and Statistics

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Sum

The probability that a machine will produce all bolts in a production run within specification is 0.998. A sample of 8 machines is taken at random. Calculate the probability that at most 6 machines will produce all bolts within specification.

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Solution

Let X = number of machines that produce the bolts within specification.

p = probability that a machine produce bolts within specification

p = 0.998

and q = 1 - p = 1 - 0.998 = 0.002

Given: n = 8

∴ X ~ B (8, 0.998)

The p.m.f. of X is given by

P(X = x) = `"^nC_x  p^x  q^(n - x)`

i.e. p(x) = `"^8C_x  (0.998)^x  (0.002)^(8 - x)`, x = 0, 1, 2,...,8

P(at most 6 machines will produce all bolts with specification)

= P[X ≤ 6] = 1 - P[x > 6]

= 1 - [P (X = 7) + P(X = 8)]

= 1 - [P(7) + P(8)]

= 1 - (1.014)(0.998)7

Hence, the probability that at most 6 machines will produce all bolts with specification = 1 - (1.014)(0.998)

Concept: Binomial Distribution
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 11.3 | Page 254
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