Question

A manufacturing firm produces a product that has a ceramic coating. The coating is baked on to the​ product, and the baking process is known to produce 20​% defective items. Every​ hour, 20 products from the thousands that are baked hourly are sampled from the​ ceramic-coating process and inspected. Complete parts a through c.

a. What is the probability that 5 defective items will be found in the next sample of 20​?

Answer 1

Given,

Random sample (n) = 20

Defective items (p) = 20% or 0.2

the probability mass function is as follows

P(x) = ⁿC_x X p^x X (1 – P)^n-x

a) The calculation of the probability that 5 defective items will be found in the next sample of 20:

P(x) = ⁿC_x X p^x X (1 – P)^n-x

P(5) = ²⁰C₅ X 0.2⁵ X (1 – 0.2)^{20 – 5}

= [20! / 5! (20 – 5)!] X 0.00032 X (0.8)¹⁵

= [20! / 5! (15)!] X 0.00032 X 0.0351844

= [(20 x 19 x18 x 17 x 16 x 15!) / 5!(15)!] X 0.00032 X 0.0351844

= (20 x 19 x18 x 17 x 16 / 5 x 4 x 3 x 2 x 1) X0.00032 X 0.0351844

= 1860480 / 120 X 0.00032 X 0.0351844

= 15504 X 0.00032 X 0.0351844

= 4.96128 X 0.0351844

= 0.17455966

= 0.1746

So, the probability that 5 defective items will be found is 0.1746

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