Respuesta :
Answer:
(a) [tex]E(X) = 950[/tex]
(b) [tex]$ COV = 0.007255$[/tex]
(c) [tex]P(X > 980) = 0.00001\\\\[/tex]
Step-by-step explanation:
The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.
probability of failure = q = 0.05
probability of success = p = 1 - 0.05 = 0.95
number of trials = n = 1000
(a) What is the expected number of non-defective units?
The expected number of non-defective units is given by
[tex]E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950[/tex]
(b) what is the COV of the number of non-defective units?
The coefficient of variance is given by
[tex]$ COV = \frac{\sigma}{E(X)} $[/tex]
Where the standard deviation is given by
[tex]\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892[/tex]
So the coefficient of variance is
[tex]$ COV = \frac{6.892}{950} $[/tex]
[tex]$ COV = 0.007255$[/tex]
(c) What is the probability of having more than 980 non-defective units?
We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.
[tex]P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\[/tex]
We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).
[tex]P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\[/tex]
The z-score corresponding to 4.28 is 0.99999
[tex]P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\[/tex]
So it means that it is very unlikely that there will be more than 980 non-defective units.
